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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If 4 marbles are picked up at random, what is the probability that at least one of them is blue?
Answer & Solution
Correct Answer:
Option
B
Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 4 marbles such that none is blue.
Then, n(E) = number of ways of drawing 4 marbles out of 11 non-blue.
$${}^{11}\mathop C\nolimits_4 = \frac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}}$$ = 330
And n(S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{330}}{{1365}}$$ $$ = \frac{{22}}{{91}}$$
∴ Required probality =$$\left( {1 - \frac{{22}}{{91}}} \right)$$ $$ = \frac{{69}}{{91}}$$
Let E be the event of drawing 4 marbles such that none is blue.
Then, n(E) = number of ways of drawing 4 marbles out of 11 non-blue.
$${}^{11}\mathop C\nolimits_4 = \frac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}}$$ = 330
And n(S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{330}}{{1365}}$$ $$ = \frac{{22}}{{91}}$$
∴ Required probality =$$\left( {1 - \frac{{22}}{{91}}} \right)$$ $$ = \frac{{69}}{{91}}$$
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Loginat least one blue=1-22/91=69/91