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A two-digit locker combination has two nonzero digits and no two digits in a combination are the same. Event A = the first digit is a prime number Event B = the second digit is a prime number If a combination is picked at random with each possible locker combination being equally likely, what is P(B|A) expressed in simplest form? A.3/8B .1/2C .4/9D. 5/9

User David Waters
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2 Answers

15 votes
15 votes

Answer:

The answer is A, 3/8

Explanation:

User Kaushik Acharya
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18 votes
18 votes

Answer:

A 3/8

Explanation:

2 digits with 9 options (as there is no 0).

and no double digits allowed.

that gives us 9 options for the first and then only 8 options for the second digit (as whatever was picked for the first, cannot be picked again for the second position).

in total

9×8 = 72

possible combinations.

and we are looking for the probability that the second digit is a prime number, when the first digit was already a prime number.

P(B|A) = P(B and A)/P(A)

we need prime numbers. what prime numbers are possible ?

2, 3, 5, 7

that means we have 4 desired cases out of 9 for theoretically each position (4/9), but again one less for the second position : 3 out of 8 (3/8).

that gives us as combined probability of A and B

4/9 × 3/8 = 12/72

the probability that the first digit is a prime number is again 4/9.

so, formally, we have for

P(B|A) = 12/72 / 4/9 = 12×9 / 4×72 = 12 / 4×8 =

= 3/8

User Ofeargall
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