In a collection of tentickets, there are two | vedclass

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Question

(c)

$p _1=\frac{{ }^2 C _1 \cdot{ }^8 C _4}{{ }^{10} C _5}=\frac{2 	imes 8 	imes 7 	imes 6 	imes 5}{24 	imes \frac{10 	imes 9 	imes 8 	imes

Vedclass · Accepted Answer

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

Vedclass · Answer

(c)

$p _1=\frac{{ }^2 C _1 \cdot{ }^8 C _4}{{ }^{10} C _5}=\frac{2 	imes 8 	imes 7 	imes 6 	imes 5}{24 	imes \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{120}}=\frac{5}{9}$

$p _2=\frac{{ }^2 C _2 \cdot{ }^8 C _3}{{ }^{10} C _5}=\frac{8 	imes 7 	imes 6}{6 	imes \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{120}}=\frac{2}{9}$

$p _1+ p _2=\frac{7}{9}$

In a collection of tentickets, there are two winning tickets. From this collection, five tickets are drawn at random Let $p_1$ and $p_2$ be the probabilities of obtaining one and two winning tickets, respectively. Then $p_1+p_2$ lies in the interval

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