3. A study by the New York Liberty women’s basketball team showed that 50% of ba

Question

3. A study by the New York Liberty women’s basketball team showed that 50% of basketball fans attending a New York Liberty basketball game will return for future games. Suppose six basketball fans are selected at random, what is the probability that: (a) Exactly two basketball fans will return? (b) All six basketball fans will return? (c) At least five basketball fans will return? (d) At least one basketball fan will return? (e) How many basketball fans would be expected to return for future Liberty games?

Explanation / Answer

X ~ Binomial(n,p)

Where n = 6 , p = 0.50

Binomial probability distribution is

P(X) = nCx px (1-p)n-x

a)

P( X = 2) = 6C2 0.502 0.504

= 0.2344

b)

P( X = 6) = 6C6 0.506 0.500

= 0.0156

c)

P( X >= 5) = P( X = 5) + P( X = 6)

= 6C5 0.505 0.50 + 6C6 0.506 0.500

= 0.1094

d)

P( X >= 1) = 1 - P( X = 0)

= 1 - 6C0 0.500 0.506

= 0.9844

e)

E(X) = n * p

= 6 * 0.50

= 3

Number of basketball fans would be expected to return for future Liberty games = 3

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.