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Two aeroplanes I and II bomb a target in succession. The probability 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 ?

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1 votes
Probability that the target is hit by the second plane means Target Hit and that bcoz of plane 2.

Target hit = 1- 0.7*0.8 = 1-0.56 = 0.44

Hit by 2nd plane = 0.2*0.7 = 0.14

So target hit by 2nd plane P = 0.14/0.44 = 0.318
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simply

0.2*0.7= 0.14

hit by 2nd plane .
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Assumption : "Since it is not explicitly given that the bomb hits the target, we have to consider both hit and miss cases." Given answer might be wrong, since probability 0.2 implies that the first plane misses the target every time, i.e. probability of missing target by plane I is 1. Consider 100 bomb drops, out of those 100 drops plane I will hit the target approximately 30 times, so plane II will get the chance to drop the bomb only 70 times out of 100 drops, now once it has get the chance to drop the bomb it will hit the target at a probability of 0.2 i.e. out of 70 times it will probably hit the target 70 x 0.2 = 14 times. So out of 100 bomb drops, plane II will probably hit the target 14 times. So possibly the answer is 0.14.
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P(H1) =0.3 ,P(H2)=0.2. So P(M1)=0.7 and P(M2)=0.8. H1 -->Hit by 1st plane,H2-->Hit by 2nd plane. M1=Miss by1st ,M2=Miss by Second

Since its a continuous process.

So the P(Hit by 2nd Plane is)=P(M1)*P(H2) +P(M1)*P(M2)*P(M1)*P(H2)+.......in this way.

                                            =P(M1)*P(H2) + P(M1)^2*P(M2)*P(H2)+....

                                          =P(H2)[P(M1)+P(M1)^2*P(M2)+P(M1)^3*P(M2)^2+.....)=

                                        =P(H2)[1/(1-P(M1)*P(M2))]= 0.2/(1-0.7*0.8)=0.454

So the answer will be 0.454.

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