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?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 = ¹⁶C4 = (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) = ²C1×⁴C2×⁶C1= 2 × 6 × 6= 72 ∴ P (E) = (n(E))/(n(S))= 72/1820 = 18/455
3 6 18 149 602 15057
...1, 1, 3, 15, 105, 942
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8, 14, 24, 49, 59, 84 2, 4, 16, 48, 240, 1440, 10080
1001, 934, 863, 790, 711, 630, 539
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1728, 121, 1000, 729, 512, 49
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21, 31, 11, 41, 1, 61Find the wrong number in the given number series.
548, 532, 568, 510, 604, 460
9 17 53 103 310 619
...43, 42, 41, 39, 33, 10
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