Question

assume that 11% of people are left handed, if we seclect 5 people at random, find the probability there are exactly 3 lefties in the group b) there are at least 3 lefties in the group c) there are no more than lefties in the group I have to show the formula used and steps Thank you

Answer 1

Solution

Given that ,

p = 11% = 0.11

1 - p = 1 - 0.11 = 0.89

n = 5

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

a)

x = 3

P(X = 3) = ((5! / 3! (5 - 3)!) * 0.11³ * (0.89)^{5 - 3}

=  ((5! / 3! (2)!) * 0.11³ * (0.89)²

= 0.0105

Probability = 0.0105

(b)

P(x 3) = 1 - P(x < 3)

= 1 - P(x = 0) - P(x = 1) - P(x = 2)

= 1 -  ((5! / 0! (5)!) * 0.11⁰ * (0.89)⁵ - ((5! / 1! (4)!) * 0.11¹ * (0.89)⁴ - ((5! / 2! (3)!) * 0.11² * (0.89)³

= 1 - 0.9888

= 0.0112

Probability = 0.0112

(c)

P(x 0) = P(x = 0) = ((5! / 0! (5)!) * 0.11⁰ * (0.89)⁵ = 0.5584

Probability = 0.5584