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 ?
A. $$\frac{7}{12}$$
B. $$\frac{37}{44}$$
C. $$\frac{5}{12}$$
D. $$\frac{7}{44}$$
E. None of these
Answer: Option B
Solution(By Examveda Team)
Total number of marbles = (4 + 5 + 3) = 12Let 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)
= $$\left( {1 - \frac{7}{{44}}} \right)$$
= $$ \frac{{37}}{{44}}$$
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
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