A student can solve 2 out of 4 problems of mathematics, 3 out of 5 problem of physics, and 4 out of 5 problems of chemistry. There are equal number of books of math, physics, and chemistry in his shelf. He selects one book randomly and attempts 10 problems from it. If he solves the first problem, then the probability that he will be able to solve the second problem is `2//3` b. `25//38` c. `13//21` d. `14//23`

Options

A. `2//3`
B. `25//38`
C. `13//21`
D. `14//23`

Answer

B

Explanation

Let `P(m),P(c),P(c)` be the probabilaty of selecting a book of maths, physics, and chemistry, respectively. Clearly,
`P(m)=P(P)=P(c)=1/3`
Again let `P(s_(1))and P(s_(2))` be the probability that he solves the first as well as second problem, respectively. Then
`P(s_(1))=P(m)xxP((s_(1))/(m))+P(P)xxP((s_(1))/(P))+P(c)xxP((s_(1))/(c))`
`impliesP(s_(1))=1/3xx1/2+1/3xx3/5+1/3xx4/5=19/30`
Similarly,
`P(S_(2))=1/3xx((1)/(2))^(2)+1/3xx((3)/(5))^(2)+1/3xx((4)/(5))^(2)=125/300`
`impliesP((S_(2))/(S_(1)))=(125/300)/(19/30)=25/38`