The probability that both the person selected at random from a group of 20 people are left handed. In the group, 15 peoples are right-handed, 4 are left-handed and 1 is ambidextrous. Two peoples are selected at random without replacement.
The probability that both the person selected at random from a group of 20 people are left handed. In the group, 15 peoples are right-handed, 4 are left-handed and 1 is ambidextrous. Two peoples are selected at random without replacement.
Solution Summary: The author explains how to calculate the probability of selecting two left-handed people from a group of 20 people.
To calculate: The probability that both the person selected at random from a group of 20 people are left handed. In the group, 15 peoples are right-handed, 4 are left-handed and 1 is ambidextrous. Two peoples are selected at random without replacement.
(b)
To determine
To calculate: The probability that one person is right-handed and one is left-handed which are selected at random from a group of 20 people. In the group, 15 peoples are right-handed, 4 are left-handed and 1 is ambidextrous. Two peoples are selected at random without replacement.
(c)
To determine
To calculate: The probability that both the peoples selected at random from a group of 20 peoples are right-handed. In the group, 15 peoples are right-handed, 4 are left-handed and 1 is ambidextrous. Two peoples are selected at random without replacement.
(d)
To determine
To calculate: The probability that both the peoples selected at random from a group of 20 peoples are ambidextrous. In the group, 15 peoples are right-handed, 4 are left-handed and 1 is ambidextrous. Two peoples are selected at random without replacement.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.
Discrete Distributions: Binomial, Poisson and Hypergeometric | Statistics for Data Science; Author: Dr. Bharatendra Rai;https://www.youtube.com/watch?v=lHhyy4JMigg;License: Standard Youtube License