Question

A bomber wants to destroy a bridge. Two bombs are sufficient to destroy it. If four bombs are dropped, what is the probability that it is destroyed, if the chances of a bomb hitting the target are 0.4?

Answer 1

Hint:
Use concept of binomial probability distribution
The binomial distribution formula helps to check the probability of getting “x” successes in “n” independent trials of a binomial experiment. The probability of failure is “1-x”. This distribution is also known as Bernoulli distribution. The performance of a fixed number of trails with fixed probability of success on each trial is known as Bernoulli trial.
The binomial distribution is a type of probability distribution that has two possible outcomes. These outcomes can be either a success or a failure. It comes with two parameters n and p.
The formula for the binomial probability distribution is as stated below:
$
P\left( x \right) = {}^n{C_r}.{p^r}{\left( {1 - p} \right)^{n - r}} \\
P\left( x \right) = {}^n{C_r}.{p^r}{\left( q \right)^{n - r}} \\
$
Where,
n = Total number of events
r = Total number of successful events
p = Probability of success on a single trial
q = Probability of failure
Here, it is given that two bombs are sufficient to destroy a bridge. Therefore, we will find the probability for two or more than two bombs.

Complete step by step solution:
According to the question -
Probability of hitting (P)$ = 0.4 = \dfrac{2}{5}$
Probability of not hitting $\left( {\bar P} \right) = q = \left( {1 - P} \right)$
$q = \left( {1 - \dfrac{2}{5}} \right) = \dfrac{3}{5}$
Now, by Probability Distribution
Probability of bridge destroyed = (Probability only two bombs hits the target) + (Probability three bombs hits the target) + (Probability four bombs hits the target)
We have, by Binomial Distribution
$
P(x = r) = {}^n{C_r}{p^r}{q^{n - r}} \\
P(x \geqslant 2) = {}^4{C_2}{p^2}{q^2} + {}^4{C_3}{p^3}{q^1} + {}^4{C_4}{p^4}{q^0} \\
$
$
= \dfrac{{4 \times 3}}{2} \times {\left( {\dfrac{2}{5}} \right)^2}{\left( {\dfrac{3}{5}} \right)^2} + 4 \times {\left( {\dfrac{2}{5}} \right)^3}{\left( {\dfrac{3}{5}} \right)^1} + {\left( {\dfrac{2}{5}} \right)^4} \times 1 \\
= \dfrac{{6 \times 4 \times 9}}{{25 \times 25}} + \dfrac{{4 \times 8 \times 3}}{{{{\left( 5 \right)}^4}}} + \dfrac{{{{\left( 2 \right)}^4}}}{{{{\left( 5 \right)}^4}}} \\
= \dfrac{{216 + 96 + 16}}{{{{\left( 5 \right)}^4}}} \\
= \dfrac{{328}}{{625}} \\
$
Therefore, the probability that it gets destroyed is $\dfrac{{328}}{{625}}$

Note:
Conditions of binomial probability distribution:
1.The number of observations is fixed.
2.Each observation is independent.
3.Each observation represents one of two outcomes (“success or failure”)
4.The probability of “success” p is the same for each outcome.