Question

The probability that a log on to the network is Successful is 0.87. Ten users attempt to log
on independently. Find the probability that between 6 and 8 are log on successful​

Answer 1

Answer:

Required probability =

$\binom{10}{6} {0.87}^{6} \times {0.13}^{4} + \binom{10}{7} {0.87}^{7} \times {0.13}^{3} + \binom{10}{8} {0.87}^{8} \times {0.13}^{2}$

Answer 2

Given:

The probability that a log on is successful (p)= 0.87

Total number of users attempting to log on = 10

To find:

The probability that between 6 and 8 users are logged on

Calculation:

The discrete probability disctribution of a binomial random variable is given by the formula

    $P(X=x)=n^c_x(p)^x(1-p){n-x}$

$\Rightarrow probability=P(X=6)+P(X=7)+P(X=6)$

                       $=C^{10}_6(0.87)^6(0.13)^{4} + C^{10}_7(0.87)^7(0.13)^{3} + C^{10}_8(0.87)^8(0.13)^{2}$

                       $= (0.87)^6\times (0.13)^{2}[C^{10}_6(0.13)^{2} + C^{10}_7(0.87)(0.13)+ C^{10}_8(0.87)^2]$

                       $=0.0073[(210\times 0.0169)+(120\times 0.1131)+(45\times 0.7569)]\\ \\ =0.0073[51.1815]\\ \\\approx0.37$    

Final answer:

The probability that between 6 and 8 users are logged on is 0.37