A and B alternately throw a pair of dice. A wins if he throws a sum of 5$5$ before B throws a sum of 8$8$, and B wins if he throws a sum of 8$8$ before A throws a sum of 5$5$. The probability, that A wins if A makes the first throw, is :
A.frac917$\frac{9}{17}$
B.frac919$\frac{9}{19}$
C.frac817$\frac{8}{17}$
D.frac819$\frac{8}{19}$
Solution & Explanation
### Related Formula
For alternating multi-stage games continuing indefinitely, the total probability of winning is modeled as an infinite geometric series summation:
S_infty = fraca1 - r$S_{\infty} = \frac{a}{1 - r}$
### Core Logic
First, analyze the total sample outcomes for a pair of standard dice (n(S) = 36$n(S) = 36$):
- Outcomes giving a sum of 5: (1,4), (2,3), (3,2), (4,1) implies 4$(1,4), (2,3), (3,2), (4,1) \implies 4$ outcomes.
p(A) = frac436 = frac19 implies p(A') = 1 - frac19 = frac89$p(A) = \frac{4}{36} = \frac{1}{9} \implies p(A') = 1 - \frac{1}{9} = \frac{8}{9}$
- Outcomes giving a sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) implies 5$(2,6), (3,5), (4,4), (5,3), (6,2) \implies 5$ outcomes.
p(B) = frac536 implies p(B') = 1 - frac536 = frac3136$p(B) = \frac{5}{36} \implies p(B') = 1 - \frac{5}{36} = \frac{31}{36}$
### Step 1: Setup Infinite Game Series Path
For player A$A$ to win on the first throw, third throw, fifth throw, etc., the probability sequence expands as follows:
P(textA wins) = p(A) + p(A') cdot p(B') cdot p(A) + [p(A') cdot p(B')]^2 cdot p(A) + dots infty$P(\text{A wins}) = p(A) + p(A') \cdot p(B') \cdot p(A) + [p(A') \cdot p(B')]^2 \cdot p(A) + \dots \infty$
This is a geometric progression where the first term a = p(A) = frac19$a = p(A) = \frac{1}{9}$ and the common ratio is:
r = p(A') cdot p(B') = frac89 cdot frac3136 = frac6281$r = p(A') \cdot p(B') = \frac{8}{9} \cdot \frac{31}{36} = \frac{62}{81}$
### Step 2: Calculate Invariant Sum
Apply the infinite GP sum formula directly:
P(textA wins) = fracfrac191 - frac6281 = fracfrac19frac1981 = frac19 cdot frac8119 = frac919$P(\text{A wins}) = \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{\frac{1}{9}}{\frac{19}{81}} = \frac{1}{9} \cdot \frac{81}{19} = \frac{9}{19}$
### Pattern Recognition
In infinite alternating games, player A's net probability can always be abbreviated shortcut-style to fracp_11 - (1-p_1)(1-p_2)$\frac{p_1}{1 - (1-p_1)(1-p_2)}$, where p_1$p_1$ is A's success probability and p_2$p_2$ is B's success probability.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Probability
Keywords:#alternating dice probability#infinite geometric progression game#JEE Main 2025 Morning Q61#sample space dice outcomes
More Probability Previous-Year Questions — Page 2
Q532025Classical Probability and Complex Powers
Two number k_1$k_1$ and k_2$k_2$ are randomly chosen from the set of natural numbers. Then, the probability that the value of i^k_1 + i^k_2$i^{k_1} + i^{k_2}$, (i = sqrt-1)$(i = \sqrt{-1})$ is non-zero, equals
(1) frac12$\frac{1}{2}$
(2) frac14$\frac{1}{4}$
(3) frac34$\frac{3}{4}$
(4) frac23$\frac{2}{3}$
A.frac12$\frac{1}{2}$
B.frac14$\frac{1}{4}$
C.frac34$\frac{3}{4}$
D.frac23$\frac{2}{3}$
Solution
### Related Formula
Probability of non-zero outcome:
P(E) = 1 - P(E^prime)$P(E) = 1 - P(E^\prime)$
where E^prime$E^\prime$ represents the condition i^k_1 + i^k_2 = 0$i^{k_1} + i^{k_2} = 0$.
### Core Logic
Powers of i$i$ repeat periodically every 4 cycles: \i, -1, -i, 1\$\{i, -1, -i, 1\}$.
Total possible outcomes for the pair (i^k_1, i^k_2)$(i^{k_1}, i^{k_2})$ are 4 times 4 = 16$4 \times 4 = 16$ cases.
### Step 1: Finding Unfavorable Cases
The sum is zero (i^k_1 + i^k_2 = 0$i^{k_1} + i^{k_2} = 0$) when the values are additive inverses:
1. (1, -1)$(1, -1)$
2. (-1, 1)$(-1, 1)$
3. (i, -i)$(i, -i)$
4. (-i, i)$(-i, i)$
This yields exactly 4 unfavorable cases.
### Step 2: Computing Probability
textProbability = frac16 - 416 = frac1216 = frac34$\text{Probability} = \frac{16 - 4}{16} = \frac{12}{16} = \frac{3}{4}$
### Pattern Recognition
Whenever modular cycles exist (like exponents of i$i$ mod 4), compress infinite natural choices safely into a single cycle map.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Complex Numbers
Class 12 Maths: Probability
Q692025Probability Distribution and Variance
Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is :
(1) 28/75
(2) 14/25
(3) 26/75
(4) 18/25
A. 28/75
B. 14/25
C. 26/75
D. 18/25
Solution
### Related Formula
Variance formula for discrete probability distributions:
sigma^2 = sum p_i x_i^2 - mu^2$\sigma^2 = \sum p_i x_i^2 - \mu^2$
### Core Logic
Construct the discrete probability distribution matrix for drawing 2 items out of 10 total items (3 defective, 7 good): Probability Distribution and Variance diagram for Q69 - JEE Main 2025 Morning
Q712025Dictionary Rank with Geometrically Weighted Probabilities
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers[cite: 666]. Let the word at serial number n be denoted by W_n$W_n$[cite: 667]. Let the probability P(W_n)$P(W_n)$ of choosing the word W_n$W_n$ satisfy P(W_n)=2P(W_n-1)$P(W_n)=2P(W_{n-1})$, n>1$n>1$[cite: 667]. If P(CDBEA)=frac2^alpha2^beta-1$P(CDBEA)=\frac{2^{alpha}}{2^{\beta}-1}$, where alpha, beta in mathbbN$\alpha, \beta \in \mathbb{N}$[cite: 687, 689], then alpha+beta$\alpha+\beta$ is equal to:
Numerical Answer.Answer: 183 to 183
Solution
### Related Formula
Sum of geometric sequence series configuration:
S_N = fraca(r^N - 1)r - 1$S_N = \frac{a(r^N - 1)}{r - 1}$
Total counts of permutations of 5 unique letter combinations = 5! = 120$= 5! = 120$.
### Core Logic
Set initial base state configuration parameter P(W_1) = x$P(W_1) = x$ [cite: 1431].
Since probabilities escalate geometrically [cite: 1433]:
sum_i=1^120 P(W_i) = x + 2x + 2^2x + dots + 2^119x = 1$\sum_{i=1}^{120} P(W_i) = x + 2x + 2^2x + \dots + 2^{119}x = 1$ [cite: 1432, 1433]
x cdot frac2^120 - 12 - 1 = 1 implies x = frac12^120 - 1$x \cdot \frac{2^{120} - 1}{2 - 1} = 1 \implies x = \frac{1}{2^{120} - 1}$ [cite: 1434]
### Step 1: Finding Dictionary Rank of CDBEA
Calculate alphabetical sequential position index values[cite: 1437]:
- Words starting with A$A$: 4! = 24$4! = 24$ [cite: 1438]
- Words starting with B$B$: 4! = 24$4! = 24$ [cite: 1438]
- Words starting with CA$CA$: 3! = 6$3! = 6$ [cite: 1438]
- Words starting with CB$CB$: 3! = 6$3! = 6$ [cite: 1439]
- Words starting with CDA$CDA$: 2! = 2$2! = 2$ [cite: 1439]
- Next word alphabetically: CDBAE$CDBAE$ (Rank 63) [cite: 1439]
- Target word: CDBEA$CDBEA$ (Rank 64) [cite: 1440]
Thus, target word index is exactly n = 64$n = 64$[cite: 1442].
### Step 2: Probabilistic coefficient calculation
Evaluate the specific targeted weighted entry value [cite: 1442]:
P(W_64) = 2^63 cdot P(W_1) = frac2^632^120-1$P(W_{64}) = 2^{63} \cdot P(W_1) = \frac{2^{63}}{2^{120}-1}$ [cite: 1442, 1443]
Matching structural values with expression flags [cite: 1443]:
alpha = 63, quad beta = 120$\alpha = 63, \quad \beta = 120$ [cite: 1443]
alpha + beta = 63 + 120 = 183$\alpha + \beta = 63 + 120 = 183$ [cite: 1443]
### Pattern Recognition
When probability states map through clean binary power scale jumps, their total summation creates standard Mersenne number forms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Class 12 Mathematics: Probability
Q732025Bayes Theorem
A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is frac1150$\frac{11}{50}$, then n is equal to
Numerical Answer.Answer: 2 to 2
Solution
### Core Logic
Let E_1$E_1$ be the event that the lost card is a spade, and E_2$E_2$ be the event that the lost card is not a spade.
P(E_1) = frac1352 = frac14 quad textand quad P(E_2) = frac3952 = frac34$P(E_1) = \frac{13}{52} = \frac{1}{4} \quad \text{and} \quad P(E_2) = \frac{39}{52} = \frac{3}{4}$
Let A$A$ be the event that n$n$ cards drawn from the remaining 51$51$ cards are all spades.
- If the lost card was a spade (E_1$E_1$), there are 12$12$ spades left out of 51$51$:
P(A|E_1) = fracbinom12nbinom51n$P(A|E_1) = \frac{\binom{12}{n}}{\binom{51}{n}}$
- If the lost card was not a spade (E_2$E_2$), there are 13$13$ spades left out of 51$51$:
P(A|E_2) = fracbinom13nbinom51n$P(A|E_2) = \frac{\binom{13}{n}}{\binom{51}{n}}$
### Step 1: Applying Bayes Theorem
We are given the posterior probability that the lost card is a spade, P(E_1|A) = frac1150$P(E_1|A) = \frac{11}{50}$:
P(E_1|A) = fracP(E_1)P(A|E_1)P(E_1)P(A|E_1) + P(E_2)P(A|E_2) = frac1150$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac{11}{50}$fracfrac14 cdot fracbinom12nbinom51nfrac14 cdot fracbinom12nbinom51n + frac34 cdot fracbinom13nbinom51n = frac1150$\frac{\frac{1}{4} \cdot \frac{\binom{12}{n}}{\binom{51}{n}}}{\frac{1}{4} \cdot \frac{\binom{12}{n}}{\binom{51}{n}} + \frac{3}{4} \cdot \frac{\binom{13}{n}}{\binom{51}{n}}} = \frac{11}{50}$
Canceling out the shared fractions frac14$\frac{1}{4}$ and binom51n$\binom{51}{n}$:
fracbinom12nbinom12n + 3binom13n = frac1150$\frac{\binom{12}{n}}{\binom{12}{n} + 3\binom{13}{n}} = \frac{11}{50}$
### Step 2: Simplifying Binomial Coefficients
Express binom13n$\binom{13}{n}$ in terms of binom12n$\binom{12}{n}$ using the identity binom13n = frac1313-nbinom12n$\binom{13}{n} = \frac{13}{13-n}\binom{12}{n}$:
fracbinom12nbinom12n + 3 cdot left[ frac1313-n binom12n right] = frac1150$\frac{\binom{12}{n}}{\binom{12}{n} + 3 \cdot \left[ \frac{13}{13-n} \binom{12}{n} \right]} = \frac{11}{50}$
Canceling binom12n$\binom{12}{n}$ from numerator and denominator:
frac11 + frac3913-n = frac1150 implies frac13-n13-n + 39 = frac1150$\frac{1}{1 + \frac{39}{13-n}} = \frac{11}{50} \implies \frac{13-n}{13-n + 39} = \frac{11}{50}$frac13-n52-n = frac1150$\frac{13-n}{52-n} = \frac{11}{50}$
Cross-multiplying:
50(13 - n) = 11(52 - n)$50(13 - n) = 11(52 - n)$650 - 50n = 572 - 11n implies 39n = 78 implies n = 2$650 - 50n = 572 - 11n \implies 39n = 78 \implies n = 2$
### Pattern Recognition
When expanding binomial dynamic ratios like fracbinomAnbinomA+1n$\frac{\binom{A}{n}}{\binom{A+1}{n}}$, always reduce the larger term fractionally using the absorption property to cancel out the factorial variables quickly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Probability
Q552025Classical Definition of Probability
The probability of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is:
A.frac129182$\frac{129}{182}$
B.frac103182$\frac{103}{182}$
C.frac1726$\frac{17}{26}$
D.frac1926$\frac{19}{26}$
Solution
### Related Formula
Classical Probability:
textP(E) = fractextFavorable CasestextTotal Cases = fracn(E)n(S)$\text{P}(E) = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{n(E)}{n(S)}$
### Core Logic
Total pool size = 4 + 2 + 10 = 16$= 4 + 2 + 10 = 16$ people. We choose 12.
Total cases n(S) = ^16mathrmC_12 = 1820$n(S) = {}^{16}\mathrm{C}_{12} = 1820$.
List distinct combinations satisfying 'at least 3 engineers' and 'at least 1 doctor':
1) 3 Engineers, 1 Doctor, 8 Professors: ^4mathrmC_3 times ^2mathrmC_1 times ^10mathrmC_8 = 4 times 2 times 45 = 360${}^{4}\mathrm{C}_{3} \times {}^{2}\mathrm{C}_{1} \times {}^{10}\mathrm{C}_{8} = 4 \times 2 \times 45 = 360$
2) 3 Engineers, 2 Doctors, 7 Professors: ^4mathrmC_3 times ^2mathrmC_2 times ^10mathrmC_7 = 4 times 1 times 120 = 480${}^{4}\mathrm{C}_{3} \times {}^{2}\mathrm{C}_{2} \times {}^{10}\mathrm{C}_{7} = 4 \times 1 \times 120 = 480$
3) 4 Engineers, 1 Doctor, 7 Professors: ^4mathrmC_4 times ^2mathrmC_1 times ^10mathrmC_7 = 1 times 2 times 120 = 240${}^{4}\mathrm{C}_{4} \times {}^{2}\mathrm{C}_{1} \times {}^{10}\mathrm{C}_{7} = 1 \times 2 \times 120 = 240$
4) 4 Engineers, 2 Doctors, 6 Professors: ^4mathrmC_4 times ^2mathrmC_2 times ^10mathrmC_6 = 1 times 1 times 210 = 210${}^{4}\mathrm{C}_{4} \times {}^{2}\mathrm{C}_{2} \times {}^{10}\mathrm{C}_{6} = 1 \times 1 \times 210 = 210$
### Step 1: Calculate Final Probability
Summing favorable cases:
n(E) = 360 + 480 + 240 + 210 = 1290$n(E) = 360 + 480 + 240 + 210 = 1290$textP(E) = frac12901820 = frac129182$\text{P}(E) = \frac{1290}{1820} = \frac{129}{182}$
### Pattern Recognition
When constraints involve 'at least' for multiple groups simultaneously with a large committee size, creating structured exhaustive case distributions is safer and less error-prone than standard bijection/subtraction methods.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Class 12 Mathematics: Probability
More Probability Questions — jee_main_2025_24_jan_morning
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