Question

if ( x ) is a binomial random variable, compute ( p(x) ) for each of the following cases:
(a) ( p(x leq 4), n = 7, p = 0.7 )
( p(x)=)
(b) ( p(x>3), n = 6, p = 0.5 )
( p(x)=0.1875 )
(c) ( p(x<5), n = 8, p = 0.6 )
( p(x)=0.4059 )
(d) ( p(x geq 7), n = 9, p = 0.4 )

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). For \(P(x\leq4)\) with \(n = 7\), \(p=0.7\), we need to calculate \(P(x = 0)+P(x = 1)+P(x = 2)+P(x = 3)+P(x = 4)\)

Step2: Calculate each term

• \(P(x = 0)=C(7,0)\times(0.7)^{0}\times(0.3)^{7}=\frac{7!}{0!(7 - 0)!}\times1\times(0.3)^{7}=1\times1\times0.0002187 = 0.0002187\)
• \(P(x = 1)=C(7,1)\times(0.7)^{1}\times(0.3)^{6}=\frac{7!}{1!(7 - 1)!}\times0.7\times(0.3)^{6}=7\times0.7\times0.000729 = 0.0035721\)
• \(P(x = 2)=C(7,2)\times(0.7)^{2}\times(0.3)^{5}=\frac{7!}{2!(7 - 2)!}\times0.49\times(0.3)^{5}=21\times0.49\times0.00243 = 0.0250047\)
• \(P(x = 3)=C(7,3)\times(0.7)^{3}\times(0.3)^{4}=\frac{7!}{3!(7 - 3)!}\times0.343\times(0.3)^{4}=35\times0.343\times0.0081 = 0.0972405\)
• \(P(x = 4)=C(7,4)\times(0.7)^{4}\times(0.3)^{3}=\frac{7!}{4!(7 - 4)!}\times0.2401\times(0.3)^{3}=35\times0.2401\times0.027 = 0.2268855\)

Step3: Sum up the terms

\(P(x\leq4)=0.0002187+ 0.0035721+0.0250047+0.0972405+0.2268855 = 0.3529215\approx0.3529\)

Answer:

\(0.3529\)