Question

One shot is fired of the three guns. E1, E2, E3 denote the events that the target is hit by the

first, second and third guns respectively. If P(E1)=0.5, P(E2)=0.6 and P(E3)=0.8 and E1, E2,E3 are

independent events, find the probability that

(a) Exactly one hit is registered

(b) At least two hits are registered​

Answer 1

Step-by-step explanation:

Given:One shot is fired of the three guns. E1, E2, E3 denote the events that the target is hit by the

first, second and third guns respectively.

If P(E1)=0.5, P(E2)=0.6 and P(E3)=0.8 and E1, E2,E3 are independent events.

To find:

find the probability that

(a) Exactly one hit is registered

(b) At least two hits are registered

Solution:

(a) Exactly one hit is registered:

1) Gun 1 hits but Gun 2 and Gun 3 not hits

2) Gun 2 hits but Gun 1 and Gun 3 not hits

3) Gun 3 hits but Gun 1 and Gun 2 not hits

P(E1)= 0.5, P(E2)=0.6, P(E3)=0.8

$P(\overline{E1})=0.5\\P(\overline{E2})=0.4\\P(\overline{E3})=0.2$

Probability of exactly one hits:

$=P(E1).P(\overline{E2}) .P(\overline{E3})+P(\overline{E1}).P(E2).P(\overline{E3})+P(\overline{E1}).P(\overline{E2}).P(E3)$

Place the values

$=0.5(0.4)(0.2)+0.5(0.6)(0.2)+0.5(0.4)(0.8)$

$=0.04+0.06+0.16$

$=0.26$

Probability of exactly one hits: 0.26

(b) At least two hits are registered:

Case 1) Any two guns hits the target

Case 2) All three guns hits the target

$=P(E1).P(E2) .P(\overline{E3})+P(\overline{E1}).P(E2).P(E3)+P(\overline{E2}).P(E1).P(E3)+P(E1)P(E2)P(E3)$

P(E1)= 0.5, P(E2)=0.6, P(E3)=0.8

Place the values

$=0.5(0.6)(0.2)+0.5(0.6)(0.8)+0.4(0.5)(0.8)+0.5(0.6)(0.8)$

$=0.06+0.24+0.16+0.24$

$=0.7$

Probability of At least two hits are registered: 0.7

Final answer:

Probability of exactly one hits: 0.26

Probability of At least two hits are registered: 0.7

Hope it helps you.

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