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
If 12 $ 18 = 10, 15 $ 18 = 11 then what is the value of 32 $ 22 = ?
If 38 × 34 = 31, 34 × 58 = 25 and 56 × 67 = 43, then 28 × 32 = ?
14 : 75 : : ?
Select the option that is related to the third number in the same way as the second number is related to first number and the sixth number is related t...
Select the option that is related to the third term in the same way as the second the term is related to the first term.
UCILVFKB : YEKNXHMF : : Q...Select the option in which the numbers are related in the same way as are the numbers in the given set.
(6, 19, 79)
If SCRIBE= HXIRYV, then FURIOUS=
E × I : 25 × 81 ∷ K × P : ?
In the given question, select the related letters from the given alternatives.
DIOT: FKLG : : UPQX : WRJC : : VSHR : ?
Study the given pattern carefully and select the number that can replace the question mark (?) in it.
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