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    Question

    A box contains 6 red, 4 blue, 2 green and 4 yellow pens.

    If four pens are picked at random, what is the probability that one is green, two are blue and one is red?
    A 13/248 Correct Answer Incorrect Answer
    B 18/455 Correct Answer Incorrect Answer
    C 14/335 Correct Answer Incorrect Answer
    D 12/545 Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    Total number of pens = 6 + 4 + 2 + 4 = 16 Let S be the sample space. Then, n(S) = number of ways of drawing 4 pens out of 16 = ¹тБ╢C4 = (16 × 15× 14 × 13 )/( 4 × 3 × 2 × 1) = 1820 Let E= event of drawing 4 pens so that one is green, two are blue and one is red = n(E) = ²C1×тБ┤C2×тБ╢C1= 2 × 6 × 6= 72 ∴ P (E) = (n(E))/(n(S))= 72/1820 = 18/455

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