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In a class of $6$ , there are $2$ students who forgot their lunch. If the teacher chooses $2$ students, what is the probability that both of them forgot their lunch?

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Probability of independent and dependent events

Full solution

Q. In a class of

6

, there are

2

students who forgot their lunch. If the teacher chooses

2

students, what is the probability that both of them forgot their lunch?

Calculate Total Ways: Determine the total number of ways to choose $2$ students from a class of $6$ . $\newline$ We use the combination formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items, and $k$ is the number of items to choose. $\newline$ For our case, $n = 6$ and $k = 2$ . $\newline$ $C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{(6\times5)}{(2\times1)} = 15$
Choose Students with No Lunch: Determine the number of ways to choose $2$ students who both forgot their lunch. $\newline$ Since there are $2$ students who forgot their lunch, we need to choose $2$ out of these $2$ . $\newline$ $C(2, 2) = \frac{2!}{(2!(2-2)!)} = \frac{2!}{(2!*0!)} = \frac{1}{1} = 1$
Calculate Probability: Calculate the probability that both students chosen forgot their lunch. $\newline$ Probability = Number of favorable outcomes / Total number of outcomes $\newline$ Probability = $\frac{C(2, 2)}{C(6, 2)} = \frac{1}{15}$

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