Conditional Probability

Question

A die is thrown 3 times and sum of 3 numbers thrown is 15. Find the chance that first thrown is 4 ?

Answer 1

Probability that first throw is 4 given that sum is 15

= P(first throw = 4, sum = 15)/ P(Sum = 15)

$= \frac{1/6 \times 2 \times 1/36}{10/216}$

$= 2/10 = 1/5.$ 

Probability that sum is 15 given that first throw is 4

= P(first throw = 4, sum = 15)/ P(First throw is 4)

$= \frac{1/6 \times 2 \times 1/36}{1/6}$

$= 2/36 = 1/18.$ 

Comments

sir i'm little confused...I think answer must be 1/5 according to the given question