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Q.

A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?

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a

5/9

b

6/3

c

3/6

d

9/5

answer is A.

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Detailed Solution

Given, the total number of red balls =6
The total number of black balls =8
The total number of white balls = 4 Total number of balls in the bag,
6+8+4= 18 Balls that are not black in bag (6 +4) = 10 As, we have 10 nonblack ball. So, overall favorable outcomes,
=

^{10} C_{1}

As, we have total 18 balls. So, total outcomes
=

^{18} C_{1}

Let the event of drawing a ball that is not black from bag =E.
Required Probability,

P (E) = \frac{(favorable outcomes)}{(Total outcomes)}

P (E) = \frac{^{10} C_{1}}{^{18} C_{1}}

P (E) = \frac{\frac{10!}{1! (10 - 1)!}}{\frac{18!}{1! (18 - 1)!}}

(since,

nCr = \frac{n!}{r! (n - r)!}

)

P (E) = \frac{10}{18}

P (E) = \frac{5}{9}

Thus,

\frac{5}{9}

is the probability that ball drawn is not black
Correct option is 1.

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