Discussion :: Probability
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A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
A.
\(\frac { 1 } { 22 } \) |
B.
\(\frac { 3 } { 22 } \) |
C.
\(\frac { 2 } { 21 }\) |
D.
\(\frac { 2 } {77 } \) |
Answer : Option C
Explanation :
Let S be the sample space.
Then, n(S) | = number of ways of drawing 3 balls out of 15 |
= 15C3 |
= \(\frac { (15*14*13) } { (3*2*1) }\)
= 455
Let E = event of getting all the 3 red balls.
n(E) = 5C3 = 5C2 =\(\frac { (5*1) } { (2 *1)} \)=10
P(E) =\(\frac { n(E) } { n(S)} \)=\( \frac { 10 } { 455 } \)=\(\frac { 2 } { 21 }\)
10/455 is 2/91.