AP EAPCET 23-Aug-2021 Shift 2 Paper

Let A be the event that aeroplane-I hit the target correctly and B be the event that aeroplane-II hit the target correctly.
⇒P(A)=0.3 and P(B)=0.2
∴P(A′)=1−P(A)=1−0.3=0.7
where A′ , first misses the target
P(B′)=1−P(B)=1−0.2=0.8
B′ , second misses the target.
Let X be the event that target is hit by plane-II
∴P(X)=P(A′)P(B) +P(A′)P(B′)P(A′)P(B) +P(A′)P(B′)P(A′)P(B′)P(A′)P(B)+.....
=0.7×0.2+0.7×0.8×0.7×0.2 +0.7×0.8×0.7×0.8×0.7×0.2+....
=0.14[1+(0.56)+(0.56)2+....]
=0.14[

1

1−0.56

] =0.14×

1

0.44

=0.32
[∵1+0.56+(0.56)2+.... forms an infinite GP
whose sum =

a

1−r

where a is first term and r is common ratio]
∴P(X)=0.32