Question
Question: On dialling certain telephone numbers, assume that on the average, one telephone number out of five ...
On dialling certain telephone numbers, assume that on the average, one telephone number out of five is busy, ten telephone numbers are randomly selected and dialled. Find the probability that at least three of them will be busy.
Solution
Hint: In order to solve this question first write what is given to us. This will give us a clear picture of what our approach should be. Also this question is solved by using the concept of Binomial distribution.
Complete step-by-step answer:
Well first we should know about it, in probability theory and statistics, the binomial distribution with the parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments.
Another thing to know about a binomial random variable is the number of successes in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution. These are some important terms to know to solve the question.
P(Probability of being busy) = 51, Q(Probability of not busy) = 1−51 and n = 10
∴ Required probability (at least three phones are busy)
=1− (Probability maximum two phones are busy)
=1−[10C0(51)0(54)10+10C1(51)1(54)9+10C2(51)2(54)8] =1−[1×1×(54)10+10×51×(54)9+45×(51)2×(54)8]
=1−(54)8[516+58+59] =1−(54)8(2516+40+45) =1−0.1678(25101) =1−0.1678×4.04 =1−0.68=0.32
Note: In this question it should be noted that this question is of chapter binomial distribution. Also one must be aware of the formula to solve the question which is mentioned in solution along with values. By these basics you should be able to solve the question. And thus we will get our desired solution.