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Question: The probability that a man can hit a target is \(\dfrac{3}{4}\) . He tries \(5\) times. The probabil...

The probability that a man can hit a target is 34\dfrac{3}{4} . He tries 55 times. The probability that he will hit the target at least three times is
1. 291364\dfrac{291}{364}
2. 371464\dfrac{371}{464}
3. 471502\dfrac{471}{502}
4. 459512\dfrac{459}{512}

Explanation

Solution

In this problem we need to calculate the probability of hitting the target more than 33 times. In the problem they have mentioned the probability of hitting as 34\dfrac{3}{4} which is equal to pp . Now we will calculate the probability of not hitting the target which is equal to qq by using the formula p+q=1p+q=1 . After having the values of pp and qq, the probability of hitting target at least three times is given by P(X≥3)P\left( X\ge 3 \right) is calculated by using the binomial distribution formula P(X=x)=nCx(p)x(q)n−xP\left( X=x \right)={}^{n}{{C}_{x}}{{\left( p \right)}^{x}}{{\left( q \right)}^{n-x}} .

Complete step by step answer:
Given, the probability of hitting the target is 34\dfrac{3}{4}.
So the value of pp will be p=34p=\dfrac{3}{4} .
Now the probability of not hitting the target qq is calculated from the formula p+q=1p+q=1, then we will have
34+q=1 ⇒q=1−34 ⇒q=14 \begin{aligned} & \dfrac{3}{4}+q=1 \\\ & \Rightarrow q=1-\dfrac{3}{4} \\\ & \Rightarrow q=\dfrac{1}{4} \\\ \end{aligned}
In the problem they have mentioned that the man tries 55 times. So the value of nn will be n=5n=5 .
Now the probability of hitting the target at least three times is given by P(X≥3)P\left( X\ge 3 \right), then we will have
P(X≥3)=P(X=3)+P(X=4)+P(X=5)P\left( X\ge 3 \right)=P\left( X=3 \right)+P\left( X=4 \right)+P\left( X=5 \right)
From the binomial distribution we can write P(X=x)=nCx(p)x(q)n−xP\left( X=x \right)={}^{n}{{C}_{x}}{{\left( p \right)}^{x}}{{\left( q \right)}^{n-x}}and use the values we have n=5n=5, p=34p=\dfrac{3}{4} and q=14q=\dfrac{1}{4} .
P(X≥3)=5C3(34)3(14)2+5C4(34)4(14)1+5C5(34)5(14)0 ⇒P(X≥3)=270+405+2431024 ⇒P(X≥3)=459512 \begin{aligned} & P\left( X\ge 3 \right)={}^{5}{{C}_{3}}{{\left( \dfrac{3}{4} \right)}^{3}}{{\left( \dfrac{1}{4} \right)}^{2}}+{}^{5}{{C}_{4}}{{\left( \dfrac{3}{4} \right)}^{4}}{{\left( \dfrac{1}{4} \right)}^{1}}+{}^{5}{{C}_{5}}{{\left( \dfrac{3}{4} \right)}^{5}}{{\left( \dfrac{1}{4} \right)}^{0}} \\\ & \Rightarrow P\left( X\ge 3 \right)=\dfrac{270+405+243}{1024} \\\ & \Rightarrow P\left( X\ge 3 \right)=\dfrac{459}{512} \\\ \end{aligned}
So, the probability of hitting a target at least three times when a man tries 55 times with a probability of hitting the target 34\dfrac{3}{4}, is 459512\dfrac{459}{512} .
So, the correct answer is “Option 4”.

Note: In this problem they have only asked to calculate the probability of hitting the target at least three times. So we have calculated the value P(X≥3)P\left( X\ge 3 \right). If they are asked to calculate the probability of hitting the target at most three times, then we need to calculate the value of P(X≤3)P\left( X\le 3 \right) by following the above mentioned procedure.