1 Answer

Nkvnkv · Answer 1 · 2022-12-20T16:46:17+0000

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,

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

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

And, their sum is:

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

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

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

Therefore, the probability that you will get “heads” no more than once out of 3 flips is
.

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