Question

Laura and Philip each fire one shot at a target. Laura has probability 0.4 of hitting the target, and Philip has probability 0.1. The shots are independent. (i). Find the probability that the target is hit.(3 points) (ii). Find the probability that the target is hit by exactly one shot.(3 points) (iii). Given that the target was hit by exactly one shot, find the probability that Philip hit the target.(4 points)

Answer 1

Answer:

(i) 0.46, (ii)0.42, and (iii)0.143

Step-by-step explanation:

Let $p$ be the probability of hitting the target and $q$ be the probability of missing the target by Laura.

Given that $p=0.4\;\cdots (1)$

As Laura either hit or miss the target, so $p+q=1$.

$\Rightarrow q=1-p=0.6 \; \cdots (2)$

Again, let $r$ be the probability of hitting the target and $s$ be the probability of missing the target by Philip.

Here, $r=0.1\;\cdots (3)$

Similarly, $\Rightarrow s=1-r=0.9\;\cdots (4)$

(i)The probability that the target is hit means that the target is not missed by both, either of one or both hit the target.

=1-(probability of missing the target by both)

=1-$qr$ [from equation (2) and (4) ]

=$1-0.6\times 0.9$

=$1-0.54$

=0.46

(ii) The probability that the target is hit by exactly one shot means either of one hit the target.

=Hit by Laura and missed by Philip or hit by Philip and missed by Laura

$=0.4\times 0.9+0.1\times 0.6$ [from equations (1),(4) and (3),(2)]

$=0.36+0.06$

=0.42

(iii)Given that the target was hit by exactly one shot, so, the given probability is 0.42 [from (ii) part]

No, the probability that the target was hit by Philip = probability of hitting the target by Philip and missing the target by Laura

$=0.1\times 0.6$ [from equations (3) and (2)]

=0.06

So, the probability of hitting the target by Philip

$=\frac {0.06}{0.42}$

$=\frac {1}{7}$

$=0.143$ (approx)