When a missile is fired from a ship, the probability that it is intercepted is (1/3) and the probability that the missile hits the target, giver that it is not intercepted, is (3/4) . If three missiles are fired independently from the ship, then the probability that all three hit the target, is

Question

When a missile is fired from a ship, the probability that it is intercepted is (1/3) and the probability that the missile hits the target, giver that it is not intercepted, is (3/4) . If three missiles are fired independently from the ship, then the probability that all three hit the target, is (A) (1/27) (B) (3/4) (C) (1/8) (D) (3/8)

Tardigrade Admin · Accepted Answer

R: Missile is intercepted H: Missile hits the target P(R)=(1/3) ; P( barR)=(2/3) ; P(H mid barR)=(3/4) beginarrayl P(H)=P(H ∩ R )+ P ( H ∩ overline R ) = P ( R ) ⋅ P ( H mid R )+ P ( overline R ) ⋅ P ( H mid overline R ) =(1/3) ⋅(0)+(2/3) ⋅ (3/4) =(1/2) endarray Hence, P ( HHH )=(1/2) ⋅ (1/2) ⋅ (1/2)=(1/8)

$\frac{1}{27}$

$\frac{3}{4}$

$\frac{1}{8}$

$\frac{3}{8}$

Q. When a missile is fired from a ship, the probability that it is intercepted is 31​ and the probability that the missile hits the target, giver that it is not intercepted, is 43​. If three missiles are fired independently from the ship, then the probability that all three hit the target, is

271​

43​

81​

83​

R: Missile is intercepted H : Missile hits the target P(R)=31​;P(Rˉ)=32​;P(H∣Rˉ)=43​ P(H)=P(H∩R)+P(H∩R)=P(R)⋅P(H∣R)+P(R)⋅P(H∣R)=31​⋅(0)+32​⋅43​=21​​ Hence, P(HHH)=21​⋅21​⋅21​=81​

$\frac{1}{27}$

$\frac{3}{4}$

$\frac{1}{8}$

$\frac{3}{8}$