Question

With n = 13 and p = 4, how do you find P(at least 7) using a binomial probability table?

Answer 1

In the given question, we have been asked to find the probability of at least 7. In order to solve the question, we first need to know about the formula of binomial distribution i.e. $P\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}\times {{p}^{r}}\times {{\left( 1-p \right)}^{n-r}}$, where n represent the number of trials and the r represents the probability of success of an event.

Complete step by step solution:
$\Rightarrow$If the probability of success of an event = p
Then, probability of failure of an event = 1-p
$\Rightarrow$The probability of ‘r’ successes out of total trial event i.e. ‘n’
Thus, $P\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}\times {{p}^{r}}\times {{\left( 1-p \right)}^{n-r}}$
We have given that,
n = 13, p = 4
Here, P (at least 7) means success of 7 or more,
Hence, desired probability = P ($r\ge 7$)
Therefore,

10,13 \right)+P\left( 11,13 \right)+P\left( 12,13 \right)+P\left( 13,13 \right)$$ As, we know that $$\Rightarrow P\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}\times {{p}^{r}}\times {{\left( 1-p \right)}^{n-r}}$$ Putting this formula, we get $$\Rightarrow P\left( r\ge 7 \right)=\left( \dfrac{13!}{7!6!}\times {{\left( 0.4 \right)}^{7}}\times {{\left( 0.7 \right)}^{6}} \right)+\left( \dfrac{13!}{8!5!}\times {{\left( 0.4 \right)}^{8}}\times {{\left( 0.7 \right)}^{5}} \right)+\left( \dfrac{13!}{9!4!}\times {{\left( 0.4 \right)}^{9}}\times {{\left( 0.7 \right)}^{4}} \right)+$$ $$\left( \dfrac{13!}{10!3!}\times {{\left( 0.4 \right)}^{10}}\times {{\left( 0.7 \right)}^{3}} \right)+\left( \dfrac{13!}{11!2!}\times {{\left( 0.4 \right)}^{11}}\times {{\left( 0.7 \right)}^{2}} \right)+\left( \dfrac{13!}{12!1!}\times {{\left( 0.4 \right)}^{12}}\times {{\left( 0.7 \right)}^{1}} \right)+\left( \dfrac{13!}{13!0!}\times {{\left( 0.4 \right)}^{13}}\times {{\left( 0.7 \right)}^{0}} \right)$$ On simplifying, we get $$\Rightarrow P\left( r\ge 7 \right)=0.13117+0.06559+0.02429+0.00648+0.00118+0.00013+0.00001$$ Adding all the numbers, we get $$\Rightarrow P\left( r\ge 7 \right)=0.22885$$ $$\therefore P\left( atleast\ 7 \right)=0.2285$$ **Formula used:** $$\Rightarrow $$If the probability of success of an event = p Then, probability of failure of an event = 1-p $$\Rightarrow $$The probability of ‘r’ successes out of total trial event i.e. ‘n’ Thus, $$P\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}\times {{p}^{r}}\times {{\left( 1-p \right)}^{n-r}}$$ **Note:** While solving this question, we need to always remember that the sum of the probabilities of the success of an event and the failure of an event always equals to 1. i.e. $$\Rightarrow $$If the probability of success of an event = p Then, probability of failure of an event = 1-p. Students should always remember that 0! It always equals 1 while solving the answer. We should always read the question carefully.