Suppose that there is a $10 \%$ probability that any given person in a population is left-handed, and a $90 \%$ probability they are not left-handed.

Say we randomly select 2 people and categorize them as left-handed (L) or not (N). Here is a sample space showing the possible outcomes:
\begin{tabular}{|c|c|}
\hline $1^{ ext {st }}$ person & $2^{ ext {nd }}$ person \
\hline$N$ & $N$ \
\hline $\mathrm{N}$ & $\mathrm{L}$ \
\hline $\mathrm{L}$ & $N$ \
\hline L & L \
\hline
\end{tabular}

Let $X$ represent the number of left-handed people in the pair.
Which table represents the theoretical probability distribution of $X$ ?
Choose 1 answer:
(A)
\begin{tabular}{c|c|c|c}
$X$ & 0 & 1 & 2 \
\hline$P(X)$ & 0.25 & 0.75 & 0.25
\end{tabular}
(B)
\begin{tabular}{c|c|c|c}
$X$ & 0 & 1 & 2 \
\hline$P(X)$ & 0.75 & 0.20 & 0.05
\end{tabular}
(C)
\begin{tabular}{c|c|c|c}
$X$ & 0 & 1 & 2 \
\hline$P(X)$ & 0.81 & 0.18 & 0.01
\end{tabular}

Question

Suppose that there is a $10 \%$ probability that any given person in a population is left-handed, and a $90 \%$ probability they are not left-handed.

Say we randomly select 2 people and categorize them as left-handed (L) or not (N). Here is a sample space showing the possible outcomes:
\begin{tabular}{|c|c|}
\hline $1^{	ext {st }}$ person & $2^{	ext {nd }}$ person \
\hline$N$ & $N$ \
\hline $\mathrm{N}$ & $\mathrm{L}$ \
\hline $\mathrm{L}$ & $N$ \
\hline L & L \
\hline
\end{tabular}

Let $X$ represent the number of left-handed people in the pair.
Which table represents the theoretical probability distribution of $X$ ?
Choose 1 answer:
(A)
\begin{tabular}{c|c|c|c}
$X$ & 0 & 1 & 2 \
\hline$P(X)$ & 0.25 & 0.75 & 0.25
\end{tabular}
(B)
\begin{tabular}{c|c|c|c}
$X$ & 0 & 1 & 2 \
\hline$P(X)$ & 0.75 & 0.20 & 0.05
\end{tabular}
(C)
\begin{tabular}{c|c|c|c}
$X$ & 0 & 1 & 2 \
\hline$P(X)$ & 0.81 & 0.18 & 0.01
\end{tabular}

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the probability of both people being not left-handed (NN). Since the probability of one person not being left-handed is $90\%$, for two people it is $0.9 imes 0.9$. ‖ ‖ step 2 ⋮ Calculate the probability of one person being left-handed and the other not (NL or LN). Since the probability of being left-handed is $10\%$ and not being left-handed is $90\%$, for one of each it is $2 imes (0.1 imes 0.9)$. ‖ ‖ step 3 ⋮ Calculate the probability of both people being left-handed (LL). Since the probability of one person being left-handed is $10\%$, for two people it is $0.1 imes 0.1$. ‖ ‖ step 4 ⋮ Assign the calculated probabilities to the correct values of $X$. $P(X=0)$ corresponds to NN, $P(X=1)$ corresponds to NL or LN, and $P(X=2)$ corresponds to LL. ‖ ∻1 Answer∻ ***C*** ∻Key Concept∻ ⚹Theoretical Probability Distribution⚹ ∻Explanation∻ ⚹The theoretical probability distribution of a random variable is a list of the probabilities of each possible outcome. In this case, it's the distribution of the number of left-handed people in a pair.⚹◊What is the formula to calculate the theoretical probability distribution of X in this scenario involving left-handed people in a population?⍭ Generate me a similar question◊

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