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A bag contains pebbles of three different colours: red, blue, and green. The number of red pebbles is 5 and the number of blue pebbles is 6. If the probability of picking a green pebble is \(\frac{7}{18}\), find the probability of picking two pebbles such that one is blue and the other is red.

1. 3/5
2. 10/51
3. 11/51
4. 9/53

Correct answer: 10/51

Solution

Let the total number of pebbles be \(N\). Since green pebbles are \(\frac{7}{18}\) of the total, red + blue = \(5+6=11\) corresponds to \(\frac{11}{18}\) of the total, so \(N=18\). Hence green pebbles = 7. The probability of drawing one red and one blue in two draws is \(\frac{5\times 6}{\binom{18}{2}}=\frac{30}{153}=\frac{10}{51}\).

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