Multiple choice

On a toss of two dice, $A$ throws a total of 5. Then the probability that be will throw another 5 before he throws 7 is

  1. $\displaystyle \frac{1}{9}$
  2. $\displaystyle \frac{1}{6}$
  3. $\displaystyle \frac{2}{5}$
  4. $\displaystyle \frac{5}{36}$
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C Correct answer
Explanation

Probability of throwing 5 is 4/36 = 1/9. Probability of throwing 7 is 6/36 = 1/6. The probability of throwing 5 before 7 is P(5) / (P(5) + P(7)) = (1/9) / (1/9 + 1/6) = (1/9) / (5/18) = 2/5.

AI explanation

The probability of throwing a 5 is 4/36 and the probability of throwing a 7 is 6/36. The conditional probability of throwing a 5 before a 7, given that one of them occurs, is the ratio of the probability of 5 to the sum of the probabilities of 5 and 7. Using conditional probability, the calculation is (4/36) divided by (4/36 + 6/36), which simplifies to 4/10 or 2/5.

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