Answer to Question #230204 in Statistics and Probability for Imii khan

Question

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
i. Less than 4 clients
ii. Exactly 4 clients
iii. At least 5 clients, will get satisfactory accommodation

Answer 1

Let the random variable X be the number of clients approaching the house agent and get satisfactory accommodation. Then we can say that X follows a binomial distribution with p=0.75 and n=6

Given the the probability of success, p, and the number of trials, n, for a binomial distribution;

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.The probability that less than 4 clients will get satisfactory accommodation

Given the binomial distribution with p=0.75 and n=6;

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

Therefore,

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

Therefore,

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