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Question: Ten eggs drawn successfully with replacement from a lot containing 10% defective eggs. Find the prob...

Ten eggs drawn successfully with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.

Explanation

Solution

Apply the Bernoulli Probability distribution formula given by P(X=a)= nCapaqn−aP\left( X=a \right)={{\text{ }}^{n}}{{C}_{a}}{{p}^{a}}{{q}^{n-a}} to calculate the probability of getting at least one defective egg. Here, ‘n’ is the number of eggs drawn, ‘a’ is the number of defective eggs drawn, ‘p’ is the probability of getting defective eggs and ‘q’ is the probability of getting non – defective eggs. To find ‘q’, apply the formula q=1–pq = 1 – p.

Complete step-by-step solution:
We have been given that the lot contains 10% defective eggs, so the percentage of non – defective eggs will be 90%. The above case is the Bernoulli distribution case. So, applying the Bernoulli probability distribution formula to find the probability of getting X = “a” eggs defective, we get,
P(X=a)= nCapaqn−aP\left( X=a \right)={{\text{ }}^{n}}{{C}_{a}}{{p}^{a}}{{q}^{n-a}}
Here, p is the probability of getting defective eggs, q is the probability of getting non – defective eggs and n is the number of eggs drawn.
Here we have been given 10% defective eggs are present in the lot.
⇒p=10100=110\Rightarrow p=\dfrac{10}{100}=\dfrac{1}{10}
⇒q=1−p=1−110=910\Rightarrow q=1-p=1-\dfrac{1}{10}=\dfrac{9}{10}
Since 10 eggs are drawn, n=10n = 10.
Now, we have to find the probability of getting at least one defective egg.
⇒a≥1\Rightarrow a\ge 1
⇒a=1,2,3....,10\Rightarrow a=1,2,3....,10
⇒P(X=a)=P(X=1)+P(X=2)+....+P(X=10).....(i)\Rightarrow P\left( X=a \right)=P\left( X=1 \right)+P\left( X=2 \right)+....+P\left( X=10 \right).....\left( i \right)
We all know that the sum of all the probabilities is 1.
⇒P(X=0)+P(X=1)+P(X=2)+....+P(X=10)=1\Rightarrow P\left( X=0 \right)+P\left( X=1 \right)+P\left( X=2 \right)+....+P\left( X=10 \right)=1
⇒P(X=1)+P(X=2)+....+P(X=3)=1−P(X=0)\Rightarrow P\left( X=1 \right)+P\left( X=2 \right)+....+P\left( X=3 \right)=1-P\left( X=0 \right)
Therefore, substituting this value in equation (i), we get,
⇒P(X=a)=P(X≥1)=1−P(X=0)\Rightarrow P\left( X=a \right)=P\left( X\ge 1 \right)=1-P\left( X=0 \right)
⇒P(X≥1)=1−P(X=0)\Rightarrow P\left( X\ge 1 \right)=1-P\left( X=0 \right)
⇒P(X≥1)=1− nC0p0qn−0\Rightarrow P\left( X\ge 1 \right)=1-{{\text{ }}^{n}}{{C}_{0}}{{p}^{0}}{{q}^{n-0}}
So, substituting the values of n = 10, p=110p=\dfrac{1}{10} and q=910,q=\dfrac{9}{10}, we get,
⇒P(X≥1)=1− 10C0(110)0(910)10\Rightarrow P\left( X\ge 1 \right)=1-{{\text{ }}^{10}}{{C}_{0}}{{\left( \dfrac{1}{10} \right)}^{0}}{{\left( \dfrac{9}{10} \right)}^{10}}
⇒P(X≥1)=1−(0.9)10\Rightarrow P\left( X\ge 1 \right)=1-{{\left( 0.9 \right)}^{10}}
Hence, this is our answer.

Note: One may note that it will be very difficult to solve the above problem without using Bernoulli probability distribution rules. We have used p and q for denoting the probability of success and failure respectively because they are used as general notation in the formula. Always remember that ‘p’ is raised to the power ‘a’ and ‘q’ to the power (n–a).(n – a).