A bag contains 5 red balls and 3 green balls. | vedclass

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(A) The total number of balls in the bag is $5 + 3 = 8$.We need to find the probability of selecting one red ball and one green ball in two draws with

Vedclass · Accepted Answer

$\frac{15}{28}$

Vedclass · Answer

(A) The total number of balls in the bag is $5 + 3 = 8$.We need to find the probability of selecting one red ball and one green ball in two draws without replacement.This can happen in two ways: (Red then Green) or (Green then Red).$	ext{Required probability} = P(R_1 \cap G_2) + P(G_1 \cap R_2)$$= P(R_1) \cdot P(G_2 | R_1) + P(G_1) \cdot P(R_2 | G_1)$$= \left(\frac{5}{8} \cdot \frac{3}{7}ight) + \left(\frac{3}{8} \cdot \frac{5}{7}ight)$$= \frac{15}{56} + \frac{15}{56}$$= \frac{30}{56} = \frac{15}{28}$

$A$ bag contains $5$ red balls and $3$ green balls. $A$ ball is selected at random and not replaced. $A$ second ball is then selected. The probability of selecting one red ball and one green ball is

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