Question

Assume that 60% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below, rounded to four decimal places:

a. There are some lefties (≥ 1) among the 5 people.

b. There are exactly 3 lefties in the group.

c. There are at least 4 lefties in the group.

d. There are no more than 2 lefties in the group.

e. How many lefties do you expect?

f. With what standard deviation?

Answer 1

a)

Here, n = 5, p = 0.6, (1 - p) = 0.4 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 1).
P(X >= 1) = (5C1 * 0.6^1 * 0.4^4) + (5C2 * 0.6^2 * 0.4^3) + (5C3 * 0.6^3 * 0.4^2) + (5C4 * 0.6^4 * 0.4^1) + (5C5 * 0.6^5 * 0.4^0)
P(X >= 1) = 0.0768 + 0.2304 + 0.3456 + 0.2592 + 0.0778
P(X >= 1) = 0.9898

b)
Here, n = 5, p = 0.6, (1 - p) = 0.4 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 3)
P(X = 3) = 5C3 * 0.6^3 * 0.4^2
P(X = 3) = 0.3456

c)
Here, n = 5, p = 0.6, (1 - p) = 0.4 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 4).
P(X >= 4) = (5C4 * 0.6^4 * 0.4^1) + (5C5 * 0.6^5 * 0.4^0)
P(X >= 4) = 0.2592 + 0.0778
P(X >= 4) = 0.3370

d)

Here, n = 5, p = 0.6, (1 - p) = 0.4 and x =2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <2).
P(X <2) = (5C0 * 0.6^0 * 0.4^5) + (5C1 * 0.6^1 * 0.4^4)
P(X <2) = 0.0102 + 0.0768
P(X <2) = 0.0870

e)

np = 5*0.60 = 3

f)

std.dev =sqrt(npq)
= sqrt(5*0.6 * 0.3)
= 0.9487