Question

Assume that 58% 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. To find the probability that there are some lefties among the 5 people, we need to find the probability of the complement event, which is that there are no lefties among the 5 people. The probability of an individual being right-handed is 1 - 0.58 = 0.42. Therefore, the probability of none of the 5 people being left-handed is:

P(no lefties) = 0.42^5 = 0.0070 (rounded to four decimal places)

The probability of there being some lefties (≥ 1) is the complement of this:

P(some lefties) = 1 - P(no lefties) = 1 - 0.0070 = 0.9930 (rounded to four decimal places)

Therefore, the probability of there being some lefties among the 5 people is 0.9930.

b. To find the probability of there being exactly 3 lefties in the group, we can use the binomial probability formula:

P(exactly 3 lefties) = (5 choose 3) * (0.58)^3 * (0.42)^2

where (5 choose 3) = 10 is the number of ways to choose 3 people out of 5. Plugging in the values, we get:

P(exactly 3 lefties) = 10 * 0.58^3 * 0.42^2 = 0.3383 (rounded to four decimal places)

Therefore, the probability of there being exactly 3 lefties among the 5 people is 0.3383.

c. To find the probability of there being at least 4 lefties in the group, we can use the binomial probability formula again:

P(at least 4 lefties) = P(4 lefties) + P(5 lefties)

P(4 lefties) = (5 choose 4) * (0.58)^4 * (0.42)^1 = 0.2684

P(5 lefties) = (5 choose 5) * (0.58)^5 * (0.42)^0 = 0.1037

Adding these probabilities, we get:

P(at least 4 lefties) = 0.2684 + 0.1037 = 0.3721 (rounded to four decimal places)

Therefore, the probability of there being at least 4 lefties among the 5 people is 0.3721.

d. To find the probability of there being no more than 2 lefties in the group, we can use the binomial probability formula again:

P(no more than 2 lefties) = P(0 lefties) + P(1 lefty) + P(2 lefties)

P(0 lefties) = (5 choose 0) * (0.58)^0 * (0.42)^5 = 0.0022

P(1 lefty) = (5 choose 1) * (0.58)^1 * (0.42)^4 = 0.0344

P(2 lefties) = (5 choose 2) * (0.58)^2 * (0.42)^3 = 0.1866

Adding these probabilities, we get:

P(no more than 2 lefties) = 0.0022 + 0.0344 + 0.1866 = 0.2232 (rounded to four decimal places)

Therefore, the probability of there being no more than 2 lefties among the 5 people is 0.2232.

Explanation: