A box contains 10 red balls and 15 green ball | vedclass

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(B) Total number of balls $= 10 + 15 = 25$.We need to draw one red ball and one green ball in two draws.The number of ways to choose $1$ red ball from

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) Total number of balls $= 10 + 15 = 25$.We need to draw one red ball and one green ball in two draws.The number of ways to choose $1$ red ball from $10$ and $1$ green ball from $15$ is given by $10 	imes 15 = 150$.The total number of ways to choose $2$ balls from $25$ is given by $^{25}C_2 = \frac{25 	imes 24}{2} = 300$.The probability is $\frac{150}{300} = \frac{1}{2}$.Alternatively,considering the order: (Red then Green) or (Green then Red).$P(RG) + P(GR) = (\frac{10}{25} 	imes \frac{15}{24}) + (\frac{15}{25} 	imes \frac{10}{24}) = \frac{150}{600} + \frac{150}{600} = \frac{300}{600} = \frac{1}{2}$.

$A$ box contains $10$ red balls and $15$ green balls. If two balls are drawn in succession,then the probability that one is red and the other is green is:

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