Aptitude Discussion

Q. |
An anti- aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the first, second, third and fourth shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the plane is hit when all the four shots are fired? |

✖ A. |
0.6872 |

✖ B. |
0.4379 |

✔ C. |
0.6976 |

✖ D. |
None of these |

**Solution:**

Option(**C**) is correct

Required probability:

\(\begin{align*} = &(0.4\times 0.7\times 0.8\times 0.9) + (0.6\times0.3\times0.8\times0.9) + \\ &(0.6\times0.7\times0.2\times0.9) + (0.6\times0.7\times0.8\times0.1) +\\ &(0.4\times0.3\times0.8\times0.9) + (0.4\times0.7\times0.2\times0.9) + \\ & (0.4\times0.7\times0.8\times0.1) + (0.6\times0.3\times0.2\times0.9) +\\ &(0.6\times0.3\times0.8\times0.1) + (0.6\times0.7\times0.2\times0.1) + \\ & (0.4\times0.3\times0.2\times0.9) + (0.6\times0.3\times0.2\times0.1) +\\ & (0.4\times0.3\times0.8\times0.1) + (0.4\times0.7\times0.2\times0.1) +\\ & (0.4\times0.3\times0.2\times0.1) \end{align*} \)

\(\begin{align*} =&0.2016+0.1296+0.756+0.336+\\ & 0.864+0.504+0.224+0.324+ \\ &0.144+0.0084+0.0216+0.0036+\\ &0.0096+0.0056+0.002 \end{align*} \)

\(=0.6976\)

**Edit:** Thank you **Vaibhav** for providing an alternative method.

**Alternate Method:**

probability that the plane is hit when all the four shots are fired,

$P=1- \text{probability of not hitting the target}$

$=1-(0.6\times 0.7 \times 0.8 \times 0.9)$

$=1-0.3024$

$=\textbf{0.6976}$

**Edit 2:** For another alternative method, check comments by **Soma Sekhar.**

**Pranjal Mishra**

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**Soma Sekhar**

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Alternate way

0.4+(0.6*0.3)+(0.6*0.7*0.7)+(0.6*0.7*0.8*0.1)

=0.6976

0.4 + 0.6x0.3 + 0.6x0.7x"0.2" + ....

**Vaibhav**

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Alternative way

1-probability of not hitting the target

=1-(0.6* 0.7 * 0.8* 0.9)

=0.6976

Thank you Vaibhav, added your method to the solution.

**Sneha**

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good one...

**Vincent**

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The answer here is so much more complicated than it needs to be. A much simpler way is to calculate the chance of NOT hitting the airplane.

First shot miss: 0.6

Second shot miss: 0.7

Third shot miss: 0.8

Fourth shot miss: 0.9

Missing with all shots:

$(0.6)(0.7)(0.8)(0.9) = 0.3024$

Thus, 0.3024 is the chance of missing all four shots. Since hitting the plane is the only other possibility, we have $1 - 0.3024 = 0.6976$ is the chance of hitting the plane with some shot.

i guess this answer is wrong because as soon as plane is hit next shot will not be taken. So if the plane is hit in the 1st shot 2nd shot is not taken

so answer will be

in 1st shot p = .4

in 2nd shot p =(1 - .4) * (.3) = .18

in 3rd shot p = (1 - .18) *(.2) = .164

in 4th shot p =(1 - .164) * (.1) = .0836

so the final P = .4 + .18 + .164 + .0836 = .8276