Answer to Question #182255 in Statistics and Probability for Iqra Raja

Question

Three missiles are fired at a target. If the probability of hitting the target are 0.3,

0.6, and 0.5 respectively. Consider that the missiles are fired independently, what

is the probability?

i) That all the missiles hit the target?

ii) That at least one hit the target?

iii) That exactly 2 hit the target?

Q.3(a)

Answer 1

Denote given events as following.

$M_1$ = "the first missile hits the target",

$M_2$ = "the second missile hits the target",

$M_3$ = "the third missile hits the target".

Consequently, the complementary events are

$\overline{M_1}$ = "the first missile does not hit the target",

$\overline{M_2}$ = "the second missile does not hit the target",

$\overline{M_3}$ = "the third missile does not hit the target".

Note that $P(\overline{M_i})=1-P(M_i)$, $i=1,2,3$ .

The probabilities are

$P(M_1)=0.3$ , $P(M_2)=0.6$ , $P(M_3)=0.5$

$P(\overline{M_1})=0.7$ , $P(\overline{M_2})=0.4$ , $P(\overline{M_3})=0.5$

For independent events use the multiplication rule

$P(M_1M_2M_3)=P(M_1)P(M_2)P(M_3)$

i) the probability that all the missiles hit the target is $P(M_1M_2M_3)=0.3\cdot0.6\cdot0.5=0.09$

ii) the probability that at least one hit the target is

$P=1-P(\overline{M_1}\ \overline{M_2}\ \overline{M_3})=$ $1-0.7\cdot0.4\cdot0.5=1-0.14=0.86$

iii) the probability that exactly 2 hit the target is

$P(M_1M_2\overline{M_3})+P(M_1\overline{M_2}M_3)+P(\overline{M_1}M_2M_3)=$

$0.3\cdot0.6\cdot0.5+0.3\cdot0.4\cdot0.5+0.7\cdot0.6\cdot0.5=$

$0.09+0.06+0.21=0.36$