Question

Harvey, laura, and gina take turns throwing spit-wads at a target. Harvey hits the target 1/2 the time, laura hits it 1/3 of the time, and gina hits the target 1/4 of the time. Given that somebody hit the target, what is the probability that it was laura?

Answer 1

Answer:

"4 / 13"

Step-by-step explanation:

Let denote the Harvey, Laura and Gina with H ,L and G respectively.

And S denote the hitting of target

Then

We are Given that

Chance of achieve the target by Harvey = P(S | H) = 1 / 2

Chance of achieve the target  by Laura   = P(S | L) = 1 / 3

Chance of achieve the target  by  Gina    = P(S | G) = 1 / 4

If they have equal chance of attempting

Then

Chance of attempting of each one = 1 / 3

that is

P(H) = P(L) = P(G) = 1/3

Total number of probabilities =P(S)= P(H)×P(S | H)+P(L)×P(S | L)+P(G)×P(S | G )

                                                          = (1/3)(1/2) + (1/3)(1/3) + (1/3)(1/4)

                                                           =1/6 + 1/9 + 1/12 = 13/36

We know that

According to Baye's theorem of probability

probability of Laure = P(L | S) = [ P(L)×P(S | L) ] / P(S)

                                                 = [ (1/3)×(1/3) ] / (13/36)

                                                = (1/9) / (13/36)

                                                = 4 / 13

So

P( L | S ) = 4 / 13