Four persons can hit a target correctly with probabilities (1/2), (1/3), (1/4) and (1/8) respectively. if all hit at the target independently, then the probability that the target would be hit, is

Question

Four persons can hit a target correctly with probabilities (1/2), (1/3), (1/4) and (1/8) respectively. if all hit at the target independently, then the probability that the target would be hit, is (A) (25/192) (B) (1/192) (C) (25/32) (D) (7/32)

Tardigrade Admin · Accepted Answer

Let persons be A,B,C,D P(Hit) = 1 - P(none of them hits) = 1 - P( barA ∩ barB ∩ barC ∩ barD) 1-P (( barA).P( barB).P( barC).P( barD) = 1 - (1/2). (2/3). (3/4). (7/8) = (25/32)

Q. Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4} an d \frac{1}{8}$ respectively. if all hit at the target independently, then the probability that the target would be hit, is

$\frac{25}{192}$

$\frac{1}{192}$

$\frac{25}{32}$

$\frac{7}{32}$

Q. Four persons can hit a target correctly with probabilities 21​,31​,41​and81​ respectively. if all hit at the target independently, then the probability that the target would be hit, is

19225​

1921​

3225​

327​

Let persons be A,B,C,D P(Hit) = 1 - P(none of them hits) =1−P(Aˉ∩Bˉ∩Cˉ∩Dˉ) 1−P((Aˉ).P(Bˉ).P(Cˉ).P(Dˉ) =1−21​.32​.43​.87​ =3225​

Q. Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4} an d \frac{1}{8}$ respectively. if all hit at the target independently, then the probability that the target would be hit, is

$\frac{25}{192}$

$\frac{1}{192}$

$\frac{25}{32}$

$\frac{7}{32}$