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There are 10 black balls and eight red balls in an urn. If four balls are drawn without replacement what is the probability that exactly 2 blackballs are drawn? Express your answer as a fraction or decimal number rounded to four decimal places

There are 10 black balls and eight red balls in an urn. If four balls are drawn without-example-1

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Solution:

Given that there are 10 black balls and 8 red balls in an urn, this implies that


\begin{gathered} n(Total)=n(black)+n(red) \\ 18=10+8 \end{gathered}

Given that 4 balls are drawn without replacement, where exactly 2 black balls are drawn, this implies that


\begin{gathered} number\text{ of ways of selecting 2 black balls:} \\ 10C2\text{ = 45 ways} \\ number\text{ of ways selecting the remaining balls from the the 8 red balls:} \\ 8C2=\text{ 28 ways} \end{gathered}

This gives


\begin{gathered} Probability\text{ of selecting exactly 2 balck balls = }(45*28)/(18C4) \\ =(45*28)/(3060) \\ =(7)/(17) \\ =0.4118(4\text{ decimal places\rparen} \end{gathered}

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