In a collection of 10 tickets,there are 2 win | vedclass

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(C) The total number of ways to draw $5$ tickets from $10$ is $^{10}C_5 = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 \

Vedclass · Accepted Answer

$\left(\frac{3}{4}, 1\right]$

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(C) The total number of ways to draw $5$ tickets from $10$ is $^{10}C_5 = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 252$.$p_1$ is the probability of obtaining exactly $1$ winning ticket:$p_1 = \frac{^2C_1 \cdot ^8C_4}{^{10}C_5} = \frac{2 	imes 70}{252} = \frac{140}{252} = \frac{5}{9}$.$p_2$ is the probability of obtaining exactly $2$ winning tickets:$p_2 = \frac{^2C_2 \cdot ^8C_3}{^{10}C_5} = \frac{1 	imes 56}{252} = \frac{56}{252} = \frac{2}{9}$.Therefore,$p_1 + p_2 = \frac{5}{9} + \frac{2}{9} = \frac{7}{9}$.Since $\frac{3}{4} = 0.75$ and $\frac{7}{9} \approx 0.777$,the value $\frac{7}{9}$ lies in the interval $\left(\frac{3}{4}, 1ight]$.

In a collection of $10$ tickets,there are $2$ winning tickets. From this collection,$5$ tickets are drawn at random. Let $p_1$ and $p_2$ be the probabilities of obtaining $1$ and $2$ winning tickets,respectively. Then $p_1+p_2$ lies in the interval:

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