A bag contains 10 different balls. Five balls are drawn simultaneouslyand then replaced and then seven balls are drawn. The probability thatexactly three balls are common to the tow draw on is `p ,`then the value of `12 p`is ______.
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Number of ways of drawing 7 balls (second draw) = `.^(10)C_(7)`
For each set of 7 balls of the second draw, 3 must be common to the set of 5 balls of the first draw, i.e., 2 other balls can be drawn in `.^(3)C_(2)` ways.
Thus, for each set of 7 balls of the second draw, there are `.^(7)C_(3) xx .^(3)C_(2)` ways of making the first draw so that there are 3 balls common.
`therefore` Probability of getting 3 balls in common = `(.^(7)C_(3) xx .^(3)C_(2))/(.^(10)C_(7))`
`=(7)/(8)`