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A and B take turn in throwing a pair of dice. A wins if he throws a total of 5 before B throws a total of 7. If A has the first throw, the probability of his winning the game is
- a)3/7
- b)9/41
- c)4/7
- d)9/11
Correct answer is option 'A'. Can you explain this answer?
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A and B take turn in throwing a pair of dice. A wins if he throws a to...
Solution:
Let's use probability tree diagram to solve this problem.
Probability tree diagram:
A has the first throw, if the sum of the numbers on the dice is 5, A wins. If the sum of the numbers on the dice is anything other than 5, the turn goes to B.
If B gets a sum of 7 before A gets 5, B wins. If B gets anything other than 7, the turn goes back to A.
Calculation:
Let's calculate the probability of A winning the game.
P(A wins on the first throw) = 4/36 = 1/9
(There are 4 ways to get a sum of 5 on a pair of dice out of the 36 possible outcomes)
P(A wins on the second throw) = (20/36)(10/36) = 50/1296
(A gets anything other than 5 on the first throw with probability 5/9. B gets anything other than 7 on the first throw with probability 25/36. So, the probability of A winning on the second throw is (5/9)(25/36)(4/36) = 50/1296)
P(A wins on the third throw) = (20/36)(26/36)(20/36)(10/36) = 250/5832
(A gets anything other than 5 on the first two throws with probability (5/9)(20/36). B gets anything other than 7 on the first two throws with probability (25/36)(26/36). A wins on the third throw with probability (5/9)(20/36)(25/36)(4/36) = 250/5832)
P(A wins on the fourth throw) = (20/36)(26/36)(20/36)(26/36)(20/36)(10/36) = 1250/139968
(The probability of A winning on the fourth throw is (5/9)(20/36)(25/36)(20/36)(25/36)(4/36) = 1250/139968)
P(A wins) = 1/9 + 50/1296 + 250/5832 + 1250/139968 = 3/7
Therefore, the probability of A winning the game is 3/7, which is option A.
Let's use probability tree diagram to solve this problem.
Probability tree diagram:
A has the first throw, if the sum of the numbers on the dice is 5, A wins. If the sum of the numbers on the dice is anything other than 5, the turn goes to B.
If B gets a sum of 7 before A gets 5, B wins. If B gets anything other than 7, the turn goes back to A.
Calculation:
Let's calculate the probability of A winning the game.
P(A wins on the first throw) = 4/36 = 1/9
(There are 4 ways to get a sum of 5 on a pair of dice out of the 36 possible outcomes)
P(A wins on the second throw) = (20/36)(10/36) = 50/1296
(A gets anything other than 5 on the first throw with probability 5/9. B gets anything other than 7 on the first throw with probability 25/36. So, the probability of A winning on the second throw is (5/9)(25/36)(4/36) = 50/1296)
P(A wins on the third throw) = (20/36)(26/36)(20/36)(10/36) = 250/5832
(A gets anything other than 5 on the first two throws with probability (5/9)(20/36). B gets anything other than 7 on the first two throws with probability (25/36)(26/36). A wins on the third throw with probability (5/9)(20/36)(25/36)(4/36) = 250/5832)
P(A wins on the fourth throw) = (20/36)(26/36)(20/36)(26/36)(20/36)(10/36) = 1250/139968
(The probability of A winning on the fourth throw is (5/9)(20/36)(25/36)(20/36)(25/36)(4/36) = 1250/139968)
P(A wins) = 1/9 + 50/1296 + 250/5832 + 1250/139968 = 3/7
Therefore, the probability of A winning the game is 3/7, which is option A.
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A and B take turn in throwing a pair of dice. A wins if he throws a total of 5 before B throws a total of 7. If A has the first throw, the probability of his winning the game isa)3/7b)9/41c)4/7d)9/11Correct answer is option 'A'. Can you explain this answer?
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