11 or 2 or 3 or 4 or none of these[3]
Two planes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is?
0.06
0.14
0.2
0.7
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11 Answers
11Actually this is a 07 aieee question
the answer given in the book is that none of these is correct
11Its 0.14
See Plane 1 is 3 in 10
plane 2 is 2 in 10
Now out of 10, 3 will be hit by plane 1 so the 7 remain
2 in 10
? in 7
1.4
Therefore 1.4 in 10
= 0.14
11yeah i do agree with msp and neil.dhruva
solution given is
prob=(0.7)(0.2)+(0.7)(0.8)(0.7)(0.2)+.........
=0.14(1/1-0.56)
=0.32
i wonder if the question means there can be more than one try?
what do you think?
3hey targetiit user i am rong becos firstly i thot the two events that is the event of the target hit by plane A and plane B are independent.
so my soln is
A be the event that plane A misses target
B be the event that plane B missses target
so reqd probability =P(A∩B)=P(A)P(B)
=(1-.3).2
=.14
thats y i am rong
11btw
can some one provide links for official solutions or answers for aieee?(if it exists [3])