Answer:
Probability that exactly one of the eggs is cracked is 0.0198.
Step-by-step explanation:
We are given that the probability that an egg on a production line is cracked is 0.01.
Two eggs are selected at random from the production line.
The above situation can be represented through binomial distribution;
[tex]P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......[/tex]
where, n = number trials (samples) taken = 2 eggs
r = number of success = exactly one
p = probability of success which in our question is probability that
an egg on a production line is cracked i.e; p = 0.01
Let X = Number of eggs on a production line that is cracked
So, X ~ Binom(n = 2, p = 0.01)
Now, probability that exactly one of the eggs is cracked is given by = P(X = 1)
P(X = 1) = [tex]\binom{2}{1} \times 0.01^{1} \times (1-0.01)^{2-1}[/tex]
= [tex]2 \times 0.01 \times (0.99)^{1}[/tex]
= 0.0198
Therefore, the probability that exactly one of the eggs is cracked is 0.0198.
