Two bags contain ten coins each and the coins in each bag are numbered from 1 to 10, one coin is drawn at random from each bag. The probability that one of the coin has value 1, 2, 3 or 4 while the other has value 7, 8, 9 or 10 is _____________.

0.32

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Solution

The correct option is A 0.32
Probability $(P_{1})$ for taking any of number (1, 2, 3, 4) from bag 1 and (7, 8, 9, 10) from bag 2

$P_{1} = \frac{4}{10} \times \frac{4}{10} = 0.16$

Similary probability $(P_{2})$ for taking any of number (7, 8, 9, 10) from bag 1 and (1, 2, 3, 4) from bag 2

$P_{2} = \frac{4}{10} \times \frac{4}{10} = 0.16$

Overall probability to draw the coins is equal to

$P_{T} = P_{1} + P_{2} = 0.16 + 0.16 = 0.32$

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Two bags contain ten coins each and the coins in each bag are numbered from 1 to 10, one coin is drawn at random from each bag. The probability that one of the coin has value 1, 2, 3 or 4 while the other has value 7, 8, 9 or 10 is _____________. 0.32

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