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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 ?
asked in Probability by (133 points) | 860 views

5 Answers

+1 vote
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
answered by Veteran (48.3k points)
its a probability which already given , so hit probability 0.44 and between them hit of A is 0.3

so hit by B is 0.44-0.3=0.14.
Out of all the HITS 31.8% hits will be by Plane II but out of all DROPS(HITS + MISSES) 14% will be by plane II.
0 votes
simply

0.2*0.7= 0.14

hit by 2nd plane .
answered by Boss (10k points)
Answer is given as 0.2. can u explain what is ur approach?
if the 1st plane is missed then only 2nd plane will have chance to

hit , it a simultaneous process thats why miss of 1st plane * hit of 2nd plane . peace
0 votes
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.
answered by Veteran (13.4k points)
0 votes
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.

Please feel free to comment
answered by (103 points)
0 votes
hit by plane1 =A=0.3
not hit by plane 1 =A'=0.7
hit by plane 2=B=0.2
not hit by plane 2=B'=0.8
hit by d 2nd one is...A'B+(A')2B'.B+(A')3(B')2B.....
=(0.7)(0.2)(1+(0.7)(0.8).......)
=(0.14)(1/1-0.56)
=0.14/0.44
=7/22
=0.32
answered by Loyal (2.8k points)

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