Question

Two aeroplanes 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

0.3

Solution

0.14

The desired probability
= 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 + … ]
$space equals space 0.14 space open square brackets fraction numerator 1 over denominator 1 minus 0.56 end fraction close square brackets space equals space fraction numerator 0.14 over denominator 0.44 end fraction space equals space 7 over 22$

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Probability

Two aeroplanes 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

0.3

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For Daily Free Study Material Join wiredfaculty

Probability

Two aeroplanes 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 0.3

Some More Questions From Probability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two aeroplanes 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

0.3