Question

Approximately 9% of people are left-handed. If two people are selected at random, what is the probability that at least one is right-handed? (Round your answer to 4 decimal places.)

Answer 1

Probabilities are used to determine the possibility of an event.

The probability that at least one of the selected two is right-handed is 0.9919

Let

$p \to$ left-handed people

$q \to$ right-handed people

So, we have:

$p = 9\%$

The probability that at least one is right-handed is calculated as follows:

$Pr =1 - P(None)$ --- Complement rule

Where:

$P(None) \to$ The probability that none of the selected 2 is right-handed; in other words, the selected 2 are left-handed.

So, we have:

$P(None) = p \times p$

$P(None) = 9\% \times 9\%$

$P(None) = 0.0081$

Substitute $P(None) = 0.0081$ in $Pr =1 - P(None)$

$Pr = 1 - 0.0081$

$Pr = 0.9919$

Hence, the probability that at least one of the selected two is right-handed is 0.9919

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