In the old days, there was a probability of 0.8 of success in any attempt to make a telephone call.Calculate the probability of having 7 successes in 10 attempts.?

probability of having 7 successes

P(x=7)

Probability P(x)=\$n_(C_x) p^x q^(n-x)\$

From the given data

n= 10,p=0.8

Finding the value of q

q = 1-p
q = 1-0.8 ⇒ q=0.2

Substitute all the values in P(x=7)

\$10_(C_7)(0.8)^7(0.2)^(10-7)\$

Simplification P(x=7)

\$10_(C_7)(0.8)^7(0.2)^3\$

\$n_(C_x) = (n!)/((n-x)!x!)\$

\$10_(C_7)\$ = \$(10!)/((10-7)!7!)\$

Simplification P(x=7)

\$(10!)/((10-7)!7!) (0.8)^7 (0.2)^3\$

After subtraction

\$(10!)/(3!7!)(0.8)^7 (0.2)^3\$

Simplification

\$(10×9×8× \cancel7!)/(3×2×1× \cancel7!) × 0.2097 × 0.008\$

After simplification

\$120 × 0.2097 × 0.008\$

Probability of having 7 successes in 10 attempts P(x=7)

0.2013

Answer

C