Probability with Permutations & Combinations


Recall that we can find the probability that event A occurs using the following formula:


Using this formula along with our knowledge of combinations and permutations, we can answer some complex probability questions.

Answering Probability Questions Using Combinations and Permutations

Problem 1: If you toss a coin four times, what is the probability that you get exactly two heads?

To find the number of ways to get two heads from four tosses, we use the formula for combinations:

4C2 = 4! / 2!(4-2)! = 4! / 2!(2!)  =  24 / 4  =  6

For each toss, there are two possible outcomes: heads or tails. Since we have four tosses, recall that we find the total number of outcomes using the formula:

(# of outcomes per trial)number of trials = 24 =16

P(exactly two heads) = 6/16 = 3/8

Problem 2: There are 10 students in a class: 5 boys and 5 girls. If the teacher randomly selects three students, what is the probability that all three students are girls?

To find the number of ways to select three girls from five total girls, we used the formula for combinations:

5C3 = 5! / 5!(5-3)! = 5! / 5!(2!)  =  10

To find the total number of ways to select three students from ten students, we again use the formula for combinations:

10C3 = 10! / 3!(10-3)! = 10! / 3!(7!)  =  120

P(all three students are girls) = 10/120 = 1/12

Problem 3: A 4-digit PIN is randomly chosen. What is the probability that there are no repeated digits?

To have no repeated digits, all four digits need to be different. This means there are 10 choices for the first digit (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9), then 9 choices for the second digit, then 8 for the third digit, and 7 for the fourth digit. Thus, the total number of ways to have no repeated digits is 10 * 9 * 8 * 7 = 5,040. We could also use the permutation formula 10P4 = 5,040.

The total number of ways to select a 4-digit pin is 10 * 10 * 10 * 10 = 104 = 10,000 total possible PINs.

P(no repeated digits) = 5,040/10,000 = 0.504

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