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Question Number 104761 by byaw last updated on 23/Jul/20

    A box contains 5 white balls, 3 black  balls and 2 red balls of the same  size. A ball is selected at random  from the box and then replaced. A  second ball is then selected. Find  the probability of obtaining    one black ball or red ball in any  order

$$ \\ $$$$\:\:\mathrm{A}\:\mathrm{box}\:\mathrm{contains}\:\mathrm{5}\:\mathrm{white}\:\mathrm{balls},\:\mathrm{3}\:\mathrm{black} \\ $$$$\mathrm{balls}\:\mathrm{and}\:\mathrm{2}\:\mathrm{red}\:\mathrm{balls}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{size}.\:\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{box}\:\mathrm{and}\:\mathrm{then}\:\mathrm{replaced}.\:\mathrm{A} \\ $$$$\mathrm{second}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{then}\:\mathrm{selected}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{obtaining}\: \\ $$$$\:\mathrm{one}\:\mathrm{black}\:\mathrm{ball}\:\mathrm{or}\:\mathrm{red}\:\mathrm{ball}\:\mathrm{in}\:\mathrm{any} \\ $$$$\mathrm{order} \\ $$

Commented by mr W last updated on 23/Jul/20

2×(2/(10))×(3/(10))=(3/(25))

$$\mathrm{2}×\frac{\mathrm{2}}{\mathrm{10}}×\frac{\mathrm{3}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{25}} \\ $$

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