Two airplanes $I$ and $I I$ bomb a target in succession. The probabilities of $I$ and $I I$ 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

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0.14

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0.2

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0.7

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Solution

The correct option is B $0.14$
The probability of plane $I$ not hitting the target $= 1 - 0.3 = 0.7$
The probability of plane $I I$ not hitting the target $= 1 - 0.2 = 0.8$
The desired probability
$= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + . . .$
$= 0.14 [1 + (0.56) + {(0.56)}^{2} + . . .]$
$= 0.14 [\frac{1}{1 - 0.56}] = \frac{0.14}{0.44} = \frac{7}{22} = 0.32$

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Two airplanes 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

Two airplanes $I$ and $I I$ bomb a target in succession. The probabilities of $I$ and $I I$ 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