Question
A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked up at random what is the probability that at least one is blue ?
Answer: Option B
Total number of marbles = (4 + 5 + 3) = 12
Let E be the event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7\mathop C\nolimits_3 $$ $$ = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$$ = 35
And, $$n(S) = {}^{12}\mathop C\nolimits_3 $$ $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$ = 220
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{35}}{{220}} = \frac{7}{{44}}$$
∴ Required probability
= 1 - P(E)
= $$\left( {1 - \frac{7}{{44}}} \right)$$
= $$ \frac{{37}}{{44}}$$
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Total number of marbles = (4 + 5 + 3) = 12
Let E be the event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7\mathop C\nolimits_3 $$ $$ = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$$ = 35
And, $$n(S) = {}^{12}\mathop C\nolimits_3 $$ $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$ = 220
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{35}}{{220}} = \frac{7}{{44}}$$
∴ Required probability
= 1 - P(E)
= $$\left( {1 - \frac{7}{{44}}} \right)$$
= $$ \frac{{37}}{{44}}$$
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