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

  • Option 1)

    0.2

  • Option 2)

    0.7

  • Option 3)

    0.06

  • Option 4)

    0.14

 

Answers (1)

As we learnt in 

Odds in favour and Against an event -

Let S be the sample space and E be an event then odd in favour of E.

- wherein

\frac{n\left ( E \right )}{n\left ( E{}' \right )}= \frac{P\left ( E \right )}{P\left ( E{}' \right )}

and     \frac{n\left ( E{}' \right )}{n\left ( E{} \right )}= \frac{P\left ( E{}' \right )}{P\left ( E{}\right )}     for odd against E

 

 

Complementary events -

Let S be the sample space  for a random experiment and let E be the event then complement of event E is 

S-n\left ( E \right )= \left n( E{}' \right )

\therefore n\left ( E \right ) + n\left ( E{}' \right )= n\left ( S \right )

- wherein

where E' contains those element which E does not contain.

 

 Probability that first plane misses  = 1- Probability the 1st plane hits

= 1-0.3\:=0.7

 Probability that 2nd plane hits = 0.2

desired events Probability = 0.7\times 0.2

= 0.14


Option 1)

0.2

Incorrect

Option 2)

0.7

Incorrect

Option 3)

0.06

Incorrect

Option 4)

0.14

Correct

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