Question 1 — Expressing Events
For events \(A,B,C\), write the following in set notation: only \(A\) occurs; exactly one occurs; at least one occurs; at least two occur; none occurs; and all three occur.
English solution
Step 0 — Dictionary: words to set notation
Events are subsets of the sample space \(\Omega\). The basic translation rules are:
- “\(A\) occurs” \(\longleftrightarrow\) the outcome lies in \(A\).
- “\(A\) does not occur” \(\longleftrightarrow\) the outcome lies in \(A^c\).
- “and / both / together” \(\longleftrightarrow\) intersection \(\cap\).
- “or / at least one of” \(\longleftrightarrow\) union \(\cup\).
Strategy
For every part, first decide, for each of \(A,B,C\), whether it must occur, must not occur, or is unrestricted. Then write \(A\) or \(A^c\) accordingly and join the pieces with \(\cap\). If several such cases are possible, join the cases with \(\cup\).
Part-by-part solution
- Only \(A\). \(A\) occurs, \(B\) does not, \(C\) does not:
\[A\cap B^c\cap C^c.\]
- Exactly one occurs. Exactly one of the three happens. There are three mutually exclusive ways: only \(A\), only \(B\), or only \(C\). Their union is the answer:
\[(A\cap B^c\cap C^c)\cup(A^c\cap B\cap C^c)\cup(A^c\cap B^c\cap C).\]
Because the three pieces are disjoint, later on their probabilities can simply be added. - At least one occurs. This just means “\(A\) or \(B\) or \(C\)”:
\[A\cup B\cup C.\]
- At least two occur. “At least two” means exactly two or all three. Equivalently, at least one pair of events occurs together:
\[(A\cap B)\cup(A\cap C)\cup(B\cap C).\]
Note that if all three occur, all three pairwise intersections happen, so the case “all three” is automatically included. - None occurs. All three fail. By De Morgan’s law,
\[A^c\cap B^c\cap C^c=(A\cup B\cup C)^c.\]
- All three occur.
\[A\cap B\cap C.\]
Extra: exactly two occur
Although not asked, it completes the picture:
\[(A\cap B\cap C^c)\cup(A\cap B^c\cap C)\cup(A^c\cap B\cap C).\]
Sanity check with the “8 regions” of a Venn diagram
Three events split \(\Omega\) into \(2^3=8\) disjoint regions (for each event: inside or outside). These regions are grouped as follows:
- none occurs: 1 region;
- exactly one occurs: 3 regions;
- exactly two occur: 3 regions;
- all three occur: 1 region.
Total \(1+3+3+1=8\) ✓. Hence “at least one” has \(3+3+1=7\) regions (everything except “none”), and “at least two” has \(3+1=4\) regions. This matches our formulas.
Another way to think: counting variable
Let \(N\) be the number of events that occur, \(N=\mathbf 1_A+\mathbf 1_B+\mathbf 1_C\). Then “exactly one” is \(\{N=1\}\), “at least one” is \(\{N\ge 1\}\), “at least two” is \(\{N\ge 2\}\), “none” is \(\{N=0\}\), and “all three” is \(\{N=3\}\).
Common mistakes
- Writing “exactly one” as \(A\cup B\cup C\). That is “at least one”; it also contains outcomes where two or three events occur.
- Writing “only \(A\)” as just \(A\), and forgetting the complements \(B^c\) and \(C^c\).
- Writing “at least two” as \(A\cap B\cap C\). That is “all three”, which is only a part of “at least two”.
Hinglish explanation
Sabse pehle basic idea
Is question mein humein sirf English sentences ko set notation mein “translate” karna hai. Bas yaad rakho: “occurs” = event set ke andar, “does not occur” = complement (\(A^c\)), “and” = \(\cap\) aur “or” = \(\cup\).
Har part ko kaise sochna hai
Har part mein teeno events \(A, B, C\) ke liye khud se poochho: “Ye hona chahiye? Ye nahi hona chahiye? Ya koi condition nahi?” Jo hona chahiye use waise hi likho, jo nahi hona chahiye uska complement likho, aur sabko \(\cap\) se jod do.
Part-by-part
- Only \(A\): “Sirf \(A\)” ka matlab \(A\) hua, lekin \(B\) aur \(C\) nahi hue. Isliye \(A\cap B^c\cap C^c\). Agar tum sirf \(A\) likh doge to woh galat hoga, kyunki \(A\) ke andar wo outcomes bhi hain jahan \(B\) ya \(C\) bhi ho jaate hain.
- Exactly one: “Bilkul ek hi” ke teen alag-alag raste hain: sirf \(A\), ya sirf \(B\), ya sirf \(C\). “Ya” ka matlab union, isliye teenon cases ka union lete hain. Ye teenon cases ek saath ho hi nahi sakte (mutually exclusive), isliye baad mein probability nikalte waqt seedha add kar sakte hain.
- At least one: “Kam se kam ek” matlab \(A\) ya \(B\) ya \(C\) mein se koi bhi. Ye simple union hai: \(A\cup B\cup C\).
- At least two: “Kam se kam do” matlab ya to exactly do hue, ya teeno hue. Isko likhne ka aasan tarika hai: koi bhi ek pair saath mein hona chahiye, yani \((A\cap B)\cup(A\cap C)\cup(B\cap C)\). Agar teeno ho gaye to teenon pairs apne aap ban jaate hain, isliye “all three” wala case alag se likhne ki zarurat nahi.
- None: Koi bhi nahi hua, yani teeno fail: \(A^c\cap B^c\cap C^c\). De Morgan law se yehi \((A\cup B\cup C)^c\) ke barabar hai — “kam se kam ek” ka ulta “koi nahi” hota hai.
- All three: Teeno saath mein: \(A\cap B\cap C\).
Apna answer check karne ka tarika
Teen events se Venn diagram mein total \(2^3=8\) chhote regions bante hain. Inmein se 1 region “none” hai, 3 regions “exactly one”, 3 regions “exactly two” aur 1 region “all three”. Total \(1+3+3+1=8\). To “at least one” = 7 regions (sirf “none” ko chhod ke) aur “at least two” = 4 regions. Ye counting hamare formulas se match karti hai, matlab answers sahi hain.
Ek aur trick
Maan lo \(N\) = kitne events hue. To “exactly one” matlab \(N=1\), “at least one” matlab \(N\ge1\), “at least two” matlab \(N\ge2\), “none” matlab \(N=0\), “all three” matlab \(N=3\). Jab bhi confuse ho, pehle sentence ko \(N\) ke terms mein likho, phir set notation banao.
Common galtiyan
(1) “Exactly one” ko \(A\cup B\cup C\) likh dena — ye “at least one” hai. (2) “Only \(A\)” mein complements bhool jaana. (3) “At least two” ko sirf \(A\cap B\cap C\) likh dena — ye to sirf “all three” hai.
Question 2 — Random Digits
Five digits are selected independently and uniformly from \(\{0,1,\ldots,9\}\). Find the probabilities that the resulting string contains (a) no 0, (b) exactly one 0, (c) at least one 0, and (d) no repeated digit.
English solution
Sample space
The outcome is an ordered string \((d_1,d_2,d_3,d_4,d_5)\) with each \(d_i\in\{0,\dots,9\}\). Since digits can repeat, there are \(10^5=100000\) strings, all equally likely, each with probability \(\frac1{100000}\).
Equivalently, look at one position at a time. Digits are independent, and for each position
\[P(d_i=0)=\frac1{10},\qquad P(d_i\ne 0)=\frac9{10}.\]
(a) No 0
Counting method: each of the 5 positions can be any of the 9 nonzero digits, so there are \(9^5=59049\) favourable strings.
Probability method: all five positions must be nonzero, so by independence
\[P(\text{no }0)=\left(\frac9{10}\right)^5=\frac{59049}{100000}=0.59049.\]
(b) Exactly one 0
Step 1: choose which position holds the 0: \(\binom51=5\) ways. Step 2: fill the remaining four positions with nonzero digits: \(9^4=6561\) ways. So favourable strings \(=5\cdot 6561=32805\), and
\[P(\text{exactly one }0)=\binom51\cdot\frac1{10}\cdot\left(\frac9{10}\right)^4=\frac{32805}{100000}=0.32805.\]
This is a Binomial\((5,\tfrac1{10})\) probability: \(P(X=1)\) where \(X\) is the number of zeros.
(c) At least one 0
Direct counting would need the cases “1 zero, 2 zeros, …, 5 zeros”. The complement of “at least one 0” is “no 0”, which we already know:
\[P(\text{at least one }0)=1-\left(\frac9{10}\right)^5=1-0.59049=\frac{40951}{100000}=0.40951.\]
Check by adding the cases: \(P(X=1)+\dots+P(X=5)=0.32805+0.07290+0.00810+0.00045+0.00001=0.40951\) ✓.
(d) No repeated digit
All five digits must be different. Choose them position by position: 10 choices for the first digit, 9 for the second (not equal to the first), 8 for the third, 7 for the fourth and 6 for the fifth. Hence the favourable count is \(10\cdot9\cdot8\cdot7\cdot6=30240=\frac{10!}{5!}\), and
\[P(\text{no repeated digit})=\frac{10\cdot9\cdot8\cdot7\cdot6}{10^5}=\frac{30240}{100000}=0.3024.\]
Note that this probability is fairly small: with 5 independent digits from only 10 symbols, a repetition is quite likely (probability \(0.6976\)).
Final answers
- (a) \(0.59049\)
- (b) \(0.32805\)
- (c) \(0.40951\)
- (d) \(0.3024\)
Hinglish explanation
Question ko samjho
Hum 5 digits ek ke baad ek choose kar rahe hain, aur har baar 0 se 9 tak koi bhi digit aa sakta hai (pehle wala digit dobara bhi aa sakta hai). Isliye total possible strings \(10\times10\times10\times10\times10=10^5=100000\) hain, aur sab equally likely hain.
Ek position ke liye: digit 0 aane ki probability \(1/10\), aur nonzero aane ki probability \(9/10\). Kyunki draws independent hain, hum alag-alag positions ki probabilities ko multiply kar sakte hain.
(a) Koi 0 nahi
Har position par nonzero digit chahiye. Har position ke liye probability \(9/10\), aur 5 positions independent hain, to \((9/10)^5=0.59049\). Counting se bhi dekho: har position ke liye 9 choices, to \(9^5=59049\) favourable strings, aur total \(100000\).
(b) Bilkul ek 0
Isme do steps hain. Step 1: 0 kis position par hoga? 5 positions hain, to 5 tareeke (\(\binom51=5\)). Step 2: baaki 4 positions par nonzero digits chahiye, har ek ke liye 9 choices, to \(9^4=6561\). Total favourable \(=5\times6561=32805\), to probability \(0.32805\).
Yaad rakho: “position choose karna” bhoolna sabse common galti hai. Agar tum sirf \(\frac1{10}\cdot(\frac9{10})^4\) likhoge to wo sirf ek particular position (jaise pehli position par 0) ki probability hogi; hume koi bhi position chalegi, isliye 5 se multiply karna zaroori hai.
(c) Kam se kam ek 0
Isko seedha count karna lamba hai (1 zero, 2 zeros, 3 zeros, 4 zeros, 5 zeros ke cases). Isliye complement trick lo: “kam se kam ek 0” ka ulta “koi 0 nahi” hai. To \(1-0.59049=0.40951\). Check karne ke liye alag-alag cases add karke bhi dekh sakte ho: \(0.32805+0.0729+0.0081+0.00045+0.00001=0.40951\) — same answer aata hai.
Rule: jab bhi “at least one” dikhe, pehle socho — kya complement (“none”) aasan hai? Aksar haan hota hai.
(d) Koi digit repeat nahi
Pehle digit ke liye 10 choices. Doosre digit ke liye 9 (kyunki pehle wale se alag hona chahiye), teesre ke liye 8, chauthe ke liye 7, aur paanchve ke liye 6. To favourable \(=10\cdot9\cdot8\cdot7\cdot6=30240\), aur probability \(30240/100000=0.3024\).
Dhyan do: yahan draws independent hain, isliye repeat hona possible hai. “No repeat” ek special condition hai, guaranteed nahi. Sirf 30% chance hai ki saare 5 digits alag honge, matlab 70% chance hai ki kam se kam ek digit repeat hoga.
Final answers
(a) \(0.59049\), (b) \(0.32805\), (c) \(0.40951\), (d) \(0.3024\).
Question 3 — Five Cards from a Standard Deck
Five cards are drawn without replacement from a standard 52-card deck. Find the probability of (a) exactly two aces, (b) at least one ace, (c) all cards of the same suit, and (d) a full house.
English solution
Sample space
The order in which the five cards arrive is irrelevant, so an outcome is a 5-card subset (a “hand”). All hands are equally likely and
\[|\Omega|=\binom{52}{5}=2{,}598{,}960.\]
A standard deck has 4 suits (13 ranks each) and 4 cards of every rank; in particular, 4 aces and 48 non-aces.
(a) Exactly two aces
Choose 2 of the 4 aces: \(\binom42=6\). The other 3 cards must be non-aces: \(\binom{48}{3}=17296\). Therefore
\[P=\frac{\binom42\binom{48}{3}}{\binom{52}{5}}=\frac{6\cdot17296}{2598960}=\frac{103776}{2598960}\approx 0.0399.\]
(b) At least one ace
The complement is “no ace”: all 5 cards come from the 48 non-aces, in \(\binom{48}{5}=1{,}712{,}304\) ways. Hence
\[P=1-\frac{\binom{48}{5}}{\binom{52}{5}}=1-\frac{1712304}{2598960}=\frac{886656}{2598960}\approx 0.3412.\]
(c) All five cards of the same suit
Choose the suit: \(4\) ways. Choose 5 of its 13 cards: \(\binom{13}{5}=1287\). So there are \(4\cdot1287=5148\) such hands and
\[P=\frac{4\binom{13}{5}}{\binom{52}{5}}=\frac{5148}{2598960}\approx 0.00198.\]
Remark. This count includes straight flushes (40 hands, including 4 royal flushes). If a poker “flush” is meant to exclude them, subtract 40: \(5108/2598960\approx0.001965\). As the problem is stated (“all cards of the same suit”), the answer is \(5148/\binom{52}{5}\).
(d) Full house (three of one rank, two of another rank)
- Choose the rank of the triple: \(13\) ways.
- Choose which 3 of its 4 suits appear: \(\binom43=4\).
- Choose the rank of the pair from the remaining ranks: \(12\) ways.
- Choose which 2 of its 4 suits appear: \(\binom42=6\).
\[P=\frac{13\binom43\cdot12\binom42}{\binom{52}{5}}=\frac{13\cdot4\cdot12\cdot6}{2598960}=\frac{3744}{2598960}\approx 0.00144.\]
The ranks are chosen as an ordered pair (triple rank first, pair rank second), because the roles are different: “three kings and two fives” is not the same hand as “three fives and two kings”. Using \(\binom{13}{2}\) here would undercount by a factor 2.
Final answers
- (a) \(\dfrac{103776}{2598960}\approx0.0399\)
- (b) \(\dfrac{886656}{2598960}\approx0.3412\)
- (c) \(\dfrac{5148}{2598960}\approx0.00198\)
- (d) \(\dfrac{3744}{2598960}\approx0.00144\)
Hinglish explanation
Sample space ka idea
Yahan 5 cards bina replacement ke nikal rahe hain aur hume sirf ye dekhna hai ki hand mein kaunse cards hain, kis order mein aaye ye matter nahi karta. Isliye combinations use karte hain. Total possible hands \(\binom{52}{5}=2{,}598{,}960\), aur sab equally likely hain. Probability = (favourable hands) ÷ (total hands).
Deck ke facts: 4 suits, har suit mein 13 cards (ranks), har rank ke 4 cards. Aces 4 hain, non-aces \(52-4=48\).
(a) Exactly do aces
Do steps: pehle 4 aces mein se 2 choose karo, \(\binom42=6\). Phir baaki 3 cards non-aces honi chahiye (warna aces ki count 2 se zyada ho jayegi), to 48 non-aces mein se 3 choose karo, \(\binom{48}{3}=17296\). Dono independent choices hain, isliye multiply: \(6\times17296=103776\). Probability \(\approx0.0399\), yani lagbhag 4%.
(b) Kam se kam ek ace
Yahan bhi complement trick sabse aasan hai. “Kam se kam ek ace” ka ulta “ek bhi ace nahi”. Ek bhi ace nahi matlab paancho cards 48 non-aces se aaye: \(\binom{48}{5}=1{,}712{,}304\). To \(1-\frac{1712304}{2598960}=\frac{886656}{2598960}\approx0.3412\), yani lagbhag 34%.
Seedha karna ho to 1 ace, 2 aces, 3 aces, 4 aces ke saare cases add karne padte — lamba kaam. Complement se ek line mein ho gaya.
(c) Saare cards same suit ke
Pehle suit choose karo: 4 options. Phir us suit ke 13 cards mein se 5 choose karo: \(\binom{13}{5}=1287\). Total \(4\times1287=5148\) hands. Probability \(5148/2598960\approx0.00198\), yani lagbhag 0.2%.
Note: is count mein straight flush wale hands (40 hands) bhi included hain, kyunki question mein sirf “same suit” likha hai. Agar poker wala “flush” chahiye (straight flush ke bina), to 40 hata dena: \(5108\).
(d) Full house
Full house matlab ek rank ke 3 cards aur doosre rank ke 2 cards, jaise KKK55. Step by step:
- Triple ki rank choose karo: 13 options.
- Us rank ke 4 suits mein se 3 choose karo: \(\binom43=4\).
- Pair ki rank choose karo: triple wali rank chhod ke 12 options bache.
- Us rank ke 4 suits mein se 2 choose karo: \(\binom42=6\).
Multiply karo: \(13\times4\times12\times6=3744\). Probability \(3744/2598960\approx0.00144\), yani lagbhag 0.14%.
Common galti: ranks ko \(\binom{13}{2}\) se choose kar lena. Wo galat hai kyunki “triple ki rank” aur “pair ki rank” ka role alag hai. KKK55 aur 555KK do alag hands hain. Isliye pehle 13 aur phir 12 use karte hain (ordered choice).
Final answers
(a) \(\approx0.0399\), (b) \(\approx0.3412\), (c) \(\approx0.00198\), (d) \(\approx0.00144\).
Question 4 — Drawing Marbles
An urn has 5 red, 4 blue, and 3 green marbles. Two marbles are drawn. Find the probability of two red marbles with replacement and without replacement; then find the probability of one red and one blue without replacement.
English solution
Setup
Total marbles: \(5+4+3=12\). Let \(R_1\) be “first marble is red”, \(R_2\) “second marble is red”, and similarly \(B_1,B_2\) for blue.
(1) Two red — with replacement
After the first draw the marble is returned, so the urn is unchanged (5 red out of 12) and the two draws are independent:
\[P(R_1\cap R_2)=P(R_1)P(R_2)=\frac5{12}\cdot\frac5{12}=\frac{25}{144}\approx0.1736.\]
(2) Two red — without replacement
Now the draws are dependent. Use the multiplication rule \(P(R_1\cap R_2)=P(R_1)P(R_2\mid R_1)\). After a red marble is removed, 11 marbles remain, of which 4 are red:
\[P(R_1\cap R_2)=\frac5{12}\cdot\frac4{11}=\frac{20}{132}=\frac5{33}\approx0.1515.\]
Check by combinations: the two marbles form an unordered pair, so \(\dfrac{\binom52}{\binom{12}{2}}=\dfrac{10}{66}=\dfrac5{33}\) ✓.
(3) One red and one blue — without replacement
“One red and one blue” can happen in two orders: red first then blue (RB), or blue first then red (BR). These are disjoint events, so we add:
\[P=\underbrace{\frac5{12}\cdot\frac4{11}}_{RB}+\underbrace{\frac4{12}\cdot\frac5{11}}_{BR}=\frac{20}{132}+\frac{20}{132}=\frac{40}{132}=\frac{10}{33}\approx0.3030.\]
Check by combinations: choose 1 red and 1 blue: \(\dfrac{\binom51\binom41}{\binom{12}{2}}=\dfrac{20}{66}=\dfrac{10}{33}\) ✓.
Remark: effect of replacement
The probability of two reds is larger with replacement (\(\frac{25}{144}\approx0.174\)) than without (\(\frac{5}{33}\approx0.152\)), because without replacement drawing a red marble first makes red marbles slightly scarcer in the urn. For comparison, if the problem asked for one red and one blue with replacement, the answer would be \(2\cdot\frac5{12}\cdot\frac4{12}=\frac5{18}\).
Final answers
- Two red, with replacement: \(\dfrac{25}{144}\)
- Two red, without replacement: \(\dfrac{5}{33}\)
- One red and one blue, without replacement: \(\dfrac{10}{33}\)
Hinglish explanation
Setup
Urn mein total \(5+4+3=12\) marbles hain (5 red, 4 blue, 3 green). Question mein dono cases hain — with replacement (marble wapas dal do) aur without replacement (marble bahar hi rehne do). Dono mein farak samajhna hi is question ka main point hai.
(1) With replacement: do red
Pehli draw ke baad marble wapas urn mein dal diya, to urn bilkul pehle jaisa ho gaya: 12 marbles, 5 red. Isliye dusri draw pehli draw se independent hai. Har draw mein red aane ki probability \(5/12\), to dono red: \(\frac5{12}\times\frac5{12}=\frac{25}{144}\approx0.174\).
(2) Without replacement: do red
Pehli draw mein red nikla (probability \(5/12\)) aur wo bahar hi reh gaya. Ab urn mein sirf 11 marbles bache, aur unme red sirf 4. To dusri draw mein red aane ki probability \(4/11\) ho jaati hai (ye conditional probability hai, \(P(R_2\mid R_1)\)). Multiplication rule se: \(\frac5{12}\times\frac4{11}=\frac{20}{132}=\frac5{33}\approx0.152\).
Combination se check: 12 marbles mein se koi bhi 2 choose karne ke \(\binom{12}{2}=66\) tareeke, aur 5 red mein se 2 choose karne ke \(\binom52=10\). To \(10/66=5/33\) ✓. Dono methods same answer dete hain.
(3) Without replacement: ek red aur ek blue
“Ek red, ek blue” do alag orders mein ho sakta hai:
- RB: pehle red, phir blue: \(\frac5{12}\times\frac4{11}\).
- BR: pehle blue, phir red: \(\frac4{12}\times\frac5{11}\).
Ye dono cases ek saath nahi ho sakte (disjoint), isliye dono ki probabilities add hoti hain: \(\frac{20}{132}+\frac{20}{132}=\frac{40}{132}=\frac{10}{33}\approx0.303\). Aksar students sirf ek order (RB) ka answer likh dete hain aur adhaa answer aata hai; dhyan rakho.
Combination se check: 1 red (5 tareeke) × 1 blue (4 tareeke) = 20 favourable pairs, total 66 pairs, \(20/66=10/33\) ✓.
Do cases ki tulna
With replacement mein “do red” ki probability zyada (\(0.174\)) hai, without replacement mein kam (\(0.152\)). Kyun? Kyunki without replacement mein ek red nikalne ke baad urn mein red ka hissa kam ho jaata hai (5/12 se 4/11), isliye dusra red aana thoda mushkil ho jaata hai.
Final answers
With replacement two red: \(\frac{25}{144}\); without replacement two red: \(\frac5{33}\); red-blue without replacement: \(\frac{10}{33}\).
Question 5 — Five People in a Line
Five distinct people \(A,B,C,D,E\) stand in a random line. Find the probability that (a) \(A\) and \(B\) are adjacent, (b) \(A\) is before \(B\), and (c) \(A\) and \(B\) are not adjacent.
English solution
Sample space
A line is a permutation of the five people. All \(5!=120\) orders are equally likely, so probability = (favourable orders) ÷ 120.
(a) \(A\) and \(B\) adjacent
Block method. Glue \(A\) and \(B\) into a single block. The block together with \(C,D,E\) gives 4 objects, which can be arranged in \(4!=24\) ways. Inside the block, the people can appear as \(AB\) or \(BA\): 2 ways. So the number of favourable lines is \(2\cdot4!=48\):
\[P(\text{adjacent})=\frac{2\cdot4!}{5!}=\frac{48}{120}=\frac25.\]
Position method (check). Choose the positions of \(A\) and \(B\) first. There are \(5\cdot4=20\) ordered position pairs. The adjacent position pairs are \((1,2),(2,3),(3,4),(4,5)\), i.e. 4 unordered pairs, so \(4\cdot2=8\) ordered ones. Thus \(\frac8{20}=\frac25\) ✓ (the other three people can fill the remaining places in \(3!\) ways, which cancels).
(b) \(A\) before \(B\)
Symmetry argument. Take any line and swap the positions of \(A\) and \(B\). This gives a one-to-one pairing between lines where \(A\) is before \(B\) and lines where \(B\) is before \(A\). The two groups therefore have the same size, and since they cover all lines,
\[P(A\text{ before }B)=\frac12.\]
Counting check: \(\frac{5!}{2}=60\) lines out of 120.
(c) \(A\) and \(B\) not adjacent
This is the complement of part (a):
\[P(\text{not adjacent})=1-\frac25=\frac35.\]
Counting check: \(120-48=72\) lines, and \(\frac{72}{120}=\frac35\) ✓.
Final answers
- (a) \(\dfrac25\)
- (b) \(\dfrac12\)
- (c) \(\dfrac35\)
Hinglish explanation
Sample space
5 alag-alag logon ko line mein khada karne ke \(5!=5\times4\times3\times2\times1=120\) tareeke hain, aur random line hai to sab 120 orders equally likely hain. Har part mein hum favourable orders count karke 120 se divide karenge.
(a) A aur B adjacent (bagal-bagal mein)
Block trick: \(A\) aur \(B\) ko ek saath chipka do aur ek single “block” maan lo. Ab hamare paas 4 cheezein hain: block, \(C\), \(D\), \(E\). In 4 cheezon ko line mein lagane ke \(4!=24\) tareeke. Lekin block ke andar bhi 2 orders possible hain: \(AB\) ya \(BA\). To favourable \(=2\times24=48\). Probability \(48/120=2/5\).
Doosra tareeka (check): Pehle sirf \(A\) aur \(B\) ki positions dekho. Positions 1 se 5 hain; adjacent position pairs ye hain: (1,2), (2,3), (3,4), (4,5) — total 4 pairs. Har pair mein \(A\), \(B\) ke 2 orders, to 8 ordered choices. \(A\), \(B\) ki koi bhi ordered positions ka total \(5\times4=20\). Isliye \(8/20=2/5\) ✓. Dono tareeke same answer dete hain, matlab pakka sahi hai.
(b) A, B se pehle
Ye bahut seedha symmetry wala sawaal hai. Kisi bhi line mein \(A\) aur \(B\) ki jagah aapas mein badal do (swap). Jahan \(A\) pehle tha wo line ab aisi ban jaayegi jahan \(B\) pehle hai, aur ulta bhi. Yaani lines ke do barabar group ban gaye: “A pehle” aur “B pehle”. Dono group same size ke, aur mil kar saari lines banate hain. To har ek ki probability \(1/2\). Counting se: \(120/2=60\) lines.
Ye trick tab bhi kaam karti hai jab line mein 5 nahi, 100 log ho — bas jab tak sab orders equally likely ho.
(c) A aur B adjacent nahi
“Adjacent nahi” bilkul “adjacent” ka complement hai. To \(1-\frac25=\frac35\). Counting se: \(120-48=72\) lines, \(72/120=3/5\) ✓.
Final answers
(a) \(\frac25\), (b) \(\frac12\), (c) \(\frac35\).
Question 6 — Neighbouring Parking Places
Five cars park at random in five consecutive spaces. Find the probability that two specified cars park next to each other.
English solution
Model
Number the spaces \(1,2,3,4,5\) in a row. “Parking at random” means that each of the \(5!=120\) assignments of the five distinct cars to the five spaces is equally likely. Call the two specified cars \(X\) and \(Y\).
Method 1 — Block method
Treat \(X\) and \(Y\) as one block. The block and the other three cars make four objects, arranged in \(4!\) ways; inside the block \(X,Y\) can swap in 2 ways. Therefore
\[P(\text{specified cars adjacent})=\frac{2\cdot4!}{5!}=\frac{48}{120}=\boxed{\frac25}.\]
Method 2 — Position method (check)
Ignore the other cars and look only at where \(X\) and \(Y\) park. There are \(\binom52=10\) equally likely pairs of spaces for them. Exactly 4 of these pairs are neighbouring: \(\{1,2\},\{2,3\},\{3,4\},\{4,5\}\). Hence
\[P=\frac{4}{\binom52}=\frac4{10}=\frac25.\]
Method 3 — Conditional (fix one car first)
Where \(X\) parks:
- end space (1 or 5), probability \(\frac25\): \(X\) has only 1 neighbouring space, so \(Y\) lands there with probability \(\frac14\);
- middle space (2, 3 or 4), probability \(\frac35\): \(X\) has 2 neighbouring spaces, so \(Y\) lands next to it with probability \(\frac24\).
\[P=\frac25\cdot\frac14+\frac35\cdot\frac24=\frac1{10}+\frac3{10}=\frac25 ✓.\]
General result
For \(n\) cars in \(n\) consecutive spaces (in a row), the same computations give
\[P(\text{two specified cars adjacent})=\frac{2\cdot(n-1)!}{n!}=\frac{2}{n}.\]
For \(n=5\) this is \(\frac25\), the same as in Question 5(a): the two problems are the same model in different words.
Remark. If the spaces were arranged in a circle, each car would have 2 neighbours out of 4 other spots, and the probability would become \(\frac12\); but here the spaces are consecutive in a row.
Final answer
\[\boxed{\frac25}\]
Hinglish explanation
Question ka matlab
5 cars ek line mein 5 parking spaces (1, 2, 3, 4, 5) mein random tarike se park hoti hain. Do specific cars, jaise \(X\) aur \(Y\), ke bagal-bagal mein park hone ki probability chahiye. Ye bilkul Question 5 (a) jaisa hai — wahan log line mein the, yahan cars spaces mein hain. Random arrangements ka total \(5!=120\).
Method 1 — Block method
\(X\) aur \(Y\) ko ek block maan lo. Ab block + baaki 3 cars = 4 objects, jinhe arrange karne ke \(4!=24\) tareeke. Block ke andar \(XY\) ya \(YX\) ho sakta hai, to 2 tareeke. Favourable \(=2\times24=48\). Probability \(\frac{48}{120}=\frac25\).
Method 2 — Sirf X aur Y ki positions dekho
Baaki 3 cars ko ignore kar do; sirf ye dekho ki \(X\) aur \(Y\) kin 2 spaces mein hain. 5 spaces mein se 2 choose karne ke \(\binom52=10\) tareeke, sab equally likely. Inmein se bagal-bagal wale pairs: (1,2), (2,3), (3,4), (4,5) — sirf 4. To \(4/10=2/5\) ✓.
Method 3 — Ek car ko pehle fix karo
Pehle \(X\) park karti hai. Agar \(X\) end space (1 ya 5) mein aayi (probability \(2/5\)), to uske bagal mein sirf 1 space hai, aur \(Y\) baaki 4 spaces mein se us ek space par \(1/4\) probability se aayegi. Agar \(X\) beech ke spaces (2, 3, 4) mein aayi (probability \(3/5\)), to uske dono taraf space hai, aur \(Y\) ke bagal mein aane ki probability \(2/4\). Total: \(\frac25\cdot\frac14+\frac35\cdot\frac24=\frac1{10}+\frac3{10}=\frac25\) ✓. Teeno methods same answer dete hain.
General formula (yaad rakhne layak)
Agar \(n\) objects line mein random arrange hon, to do specified objects ke adjacent hone ki probability \(\frac2n\) hoti hai. Yahan \(n=5\), to \(\frac25\). Ye formula block method se aata hai: \(\frac{2\cdot(n-1)!}{n!}=\frac2n\).
Ek chhota sa remark: agar parking spaces gol (circle mein) hoti, to har car ke exactly 2 neighbours hote aur baaki 4 jagahon mein se 2 mein aana hota, to answer \(\frac12\) hota. Yahan spaces seedhi line mein hain, isliye \(\frac25\).
Final answer
\(\dfrac25\)
Quick Revision — Answers and Key Ideas
English summary
- Q1: Translate words to sets. “Only \(A\)” needs complements; “exactly one” is a union of three disjoint cases; “at least two” \(=(A\cap B)\cup(A\cap C)\cup(B\cap C)\); “none” \(=(A\cup B\cup C)^c\).
- Q2: (a) \(0.59049\), (b) \(0.32805\), (c) \(0.40951\), (d) \(0.3024\). Ideas: independence, choose the position of the 0, complement for “at least one”.
- Q3: (a) \(\frac{\binom42\binom{48}{3}}{\binom{52}{5}}\approx0.0399\), (b) \(1-\frac{\binom{48}{5}}{\binom{52}{5}}\approx0.3412\), (c) \(\frac{4\binom{13}{5}}{\binom{52}{5}}\approx0.00198\), (d) \(\frac{13\binom43\,12\binom42}{\binom{52}{5}}\approx0.00144\).
- Q4: \(\frac{25}{144}\) (with replacement), \(\frac5{33}\) (without replacement), red-blue \(\frac{10}{33}\) (add both orders).
- Q5: adjacent \(\frac25\), A before B \(\frac12\), not adjacent \(\frac35\).
- Q6: \(\frac25\) (general: \(\frac2n\)).
Hinglish summary
Exam se pehle ye yaad rakho: (1) “At least one” dikhe to complement soch lo. (2) Order matter na kare to combinations, order matter kare to permutations ya multiplication rule. (3) “Exactly / only” wale sawaalon mein pehle choose karo kaunsi position ya kaunse items, phir baaki ko fill karo. (4) Do outcomes ke alag-alag orders (jaise RB aur BR) ho sakte hain to dono ko add karo. (5) “Adjacent” ke liye block method use karo aur block ke andar ke orders (\(\times2\)) mat bhoolo. (6) Har answer ko kisi doosre tareeke se ek baar cross-check karne ki aadat daalo.