Question

3) Someone shoots 4 arrows at a target. These shots are independent of each other as events. The probabilities that they hit the target are given as follows: If three of them hit the target, compute the probability that they were the three first ones.

Answer 1

Given that P₁=0.1,P₂=0.2, P₃=0.3, P₄=0.4

Let E₁ be the event of first shot hit the target .

P(E₁)=0.1

Let E₂ be the event of second shot hit the target .

P(E₂)=0.2

Let E₃ be the event of third shot hit the target

P(E₃)=0.3

Let E₄ be the event of court shot hit the target

P(E₄)=0.4

P(E'₄)=0.6

The probability of first three shots hit the target is

P(E₁$\, \bigcap$ E_{2$\, \bigcap$}E_{3$\, \bigcap$}E'₄)= P(E₁).P(E₂).P(E₃).P(E'₄)

=0.1X 0.2 X 0.3 X0.6 =0.0036.