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An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.
Answer & Solution
Correct Answer:
Option
B
Total number of balls = (2 + 3 + 2) = 7
Let, E be the event of drawing 2 non-blue balls.
Then, n (E) = $${}^5\mathop C\nolimits_4 = \frac{{5 \times 4}}{{2 \times 1}}$$ = 10
And, n (S) = $${}^7\mathop C\nolimits_2 = \frac{{7 \times 6}}{{2 \times 1}}$$ = 21
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{10}}{{21}}$$
Let, E be the event of drawing 2 non-blue balls.
Then, n (E) = $${}^5\mathop C\nolimits_4 = \frac{{5 \times 4}}{{2 \times 1}}$$ = 10
And, n (S) = $${}^7\mathop C\nolimits_2 = \frac{{7 \times 6}}{{2 \times 1}}$$ = 21
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{10}}{{21}}$$
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Login5C2 = 5!/2!*33!= 10