Out of a group of 10 people,there are 5 lawye | vedclass

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(A) Total number of ways to select $4$ people out of $10$ is $n(S) = {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} =

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$1/2$

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(A) Total number of ways to select $4$ people out of $10$ is $n(S) = {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.To have at least one person from each profession in a group of $4$,the possible combinations of professions $(L, D, E)$ are:Case $1$: $(2, 1, 1) \implies {}^{5}C_2 	imes {}^{3}C_1 	imes {}^{2}C_1 = 10 	imes 3 	imes 2 = 60$.Case $2$: $(1, 2, 1) \implies {}^{5}C_1 	imes {}^{3}C_2 	imes {}^{2}C_1 = 5 	imes 3 	imes 2 = 30$.Case $3$: $(1, 1, 2) \implies {}^{5}C_1 	imes {}^{3}C_1 	imes {}^{2}C_2 = 5 	imes 3 	imes 1 = 15$.Total favorable outcomes $n(E) = 60 + 30 + 15 = 105$.Probability $P(E) = \frac{n(E)}{n(S)} = \frac{105}{210} = \frac{1}{2}$.

Out of a group of $10$ people,there are $5$ lawyers,$3$ doctors,and $2$ engineers. If $4$ people are selected at random,what is the probability that at least one person from each profession is included?

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