Question

The probability that a bomb hits a target is 1/4.five bombs are aimed at the target. Find the probability that
1. 3 bombs hit a target
2. At the most 2 bombs hit the target

Answer 1

Step-by-step explanation:

1. 3/5
2. 2/5

so this is the answer

Answer 2

(1) The probability that  3 bombs hit the target is 45/128

(2) the probability that  at most 2 bombs hit the target is 167/256

Step-by-step explanation:

Let an event E be

E : Bomb hits the target

Then

Probability of event E

$P(E)=\frac{1}{4}$

Therefore,

$P(\bar E)=1-\frac{1}{4}=\frac{3}{4}$

(1) Let event A be

A: 3 bombs hit the target

[tex[P(A)=^5C_3(\frac{1}{4})^3(\frac{3}{4})^2[/tex]

or $P(A)=\frac{5\times 4}{2}\times \frac{9}{4^3\times 4^2}$

or $P(A)=\frac{45}{128}$

(2) Let event B be

B: At most 2 bombs hit the target

P(B) = 1 - Probability that 3 bombs hit the target - Probability that 4 bombs hit the target

$P(B)=1-\frac{45}{128}-(\frac{1}{4})^4$

$P(B)=1-(\frac{45}{128}-\frac{1}{256})$

$P(B)=1-\frac{89}{256}$

$P(B)=\frac{167}{256}$

Hope this answer is helpful.

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