Question

What is the probability that you will get “heads” no more than once out of 3 flips?
P(0 successes) + P(1 success) =

Answer 1

Given:

Three coins are tossed.

To find:

The probability that you will get “heads” no more than once out of 3 flips.

Solution:

If three coins are tossed then the sample space is:

$S=\{HHH,HTH,HHT,HTT,THH,TTH,THT,TTT\}$

Total possible outcomes = 8

Number of 0 successes means 0 heads in 3 flips, i.e., TTT.

Number of 1 successes means exactly 1 heads in 3 flips, i.e., HTT, THT, TTH.

Now,

$P(0\text{ success})=\dfrac{1}{8}$

$P(1\text{ success})=\dfrac{3}{8}$

And, their sum is:

$P(0\text{ success})+P(1\text{ success})=\dfrac{1}{8}+\dfrac{3}{8}$

$P(0\text{ success})+P(1\text{ success})=\dfrac{4}{8}$

$P(0\text{ success})+P(1\text{ success})=\dfrac{1}{2}$

Therefore, the probability that you will get “heads” no more than once out of 3 flips is $\dfrac{1}{2}$.