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Question

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

If 3 pencils are picked at random, what is the probability that two are red and one is yellow?

A 3/28 Correct Answer Incorrect Answer
B 2/21 Correct Answer Incorrect Answer
C 4/35 Correct Answer Incorrect Answer
D 2/45 Correct Answer Incorrect Answer
E none of these Correct Answer Incorrect Answer

Solution

Total number of pencils = 6 + 4 + 2 + 4 = 16 Let S be the sample space. Then, n(S) = number of ways of drawing 3 pencils out of 16 = 16c3[if gte msEquation 12]>¹⁶ C 3 [if !msEquation]--> [endif]-->= `(16xx15xx14)/(3xx2xx1)` ` ` = 560 Let E= event of drawing 3 pencils so that two are red and one is yellow = n(E) = 6C2 x 4C1 [if gte msEquation 12]>⁶ C 2 × ⁴ C 1 [if !msEquation]--> [endif]-->= 15 `xx` 4= 60 [if gte msEquation 12]>∴ [if !msEquation]--> [endif]-->P (E) =`(n(E))/(n(S))=60/560=3/28`

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