Answer to Question #229879 in Statistics and Probability for m.hasnain

Question

Note. Apply Discrete Probability Distributions to solve the given problems

Q1. The experience of a house agent indicates that he can provide suitable accomodation for 75% of the clients who come to him. If on a particular occasion, 6 clients approach him indepndently, calculate the probability that   

1. Less than 4 clients
2. Exactly 4 clients
3. At least 5 clients, will get satisfactory accommodation 

Answer 1

Suppose we let X = the number of clients approaching the house agent who get satisfactory accommodation, then the random variable X follows a binomial distribution with n=6 and p=0.75

Given the number of trials, n, and the probability of success, p,

P(X=x) = $\begin{pmatrix} n \\ X \end{pmatrix}$p^x(1-p)^n-x for x = 0, 1,2,3,4,5 and 6

1.Less than 4 clients

Given n=6 and p=0.75

P(X<4) = P(x=0) +P(x=1) + P(x=2) +P(x=3)

P(x=0) = $\begin{pmatrix} 6 \\ 0 \end{pmatrix}$0.75⁰(1-0.75)^6-0 = 0.0002 

P(x=1) = $\begin{pmatrix} 6 \\ 1 \end{pmatrix}$0.75¹(1-0.75)^6-1 = 0.0044

P(x=2) =   $\begin{pmatrix} 6 \\ 2 \end{pmatrix}$0.75²(1-0.75)^6-2 = 0.0330

P(x=3) = $\begin{pmatrix} 6 \\ 3 \end{pmatrix}$0.75³(1-0.75)^6-3 = 0.1318

Thus,

P(X<4) = 0.0002 + 0.0044 +0.0330 + 0.1318 = 0.1694

Therefore, the probability that less than 4 clients will get satisfactory accommodation is approximately equal to 0.1694

2.Exactly 4 clients

P(4) =   $\begin{pmatrix} 6 \\ 4 \end{pmatrix}$0.75⁴(1-0.75)^6-4 = 0.2966

The probability that exactly four clients will get satisfactory accommodation is approximately equal to 0.2966

3.At least 5 clients, will get satisfactory accommodation

P(X≥5) = P(5) + P(6)

P(5) = $\begin{pmatrix} 6 \\ 5 \end{pmatrix}$0.75⁵(1-0.75)^6-5 = 0.3560

P(6) =   $\begin{pmatrix} 6 \\ 6 \end{pmatrix}$0.75⁶(1-0.75)^6-6 = 0.1780

Thus,

P(X≥5) = P(5) + P(6) = 0.3560 + 0.1780 ≃ 0.5340

Therefore, the probability that at least 5 clients will get satisfactory accommodation is approximately equal to 0.5340