Number of ways of selecting 4 marbles when no one is blue = \(\displaystyle ^{{11}}{{C}_{4}}=\frac{{11\times 10\times 9\times 8}}{{4\times 3\times 2\times 1}}=330\)
Probability of getting 4 marble (when no one is blue) = \(\displaystyle \frac{{330}}{{1365}}=\frac{{22}}{{91}}\)
Probability that at least one is blue = \(\displaystyle 1-\frac{{22}}{{91}}=\frac{{69}}{{91}}\)
2. If two marbles are picked at random, what is the probability that both are red?
(a) \(\displaystyle \frac{1}{{6}}\)
(b) \(\displaystyle \frac{1}{3}\)
(c) \(\displaystyle \frac{2}{15}\)
(d) \(\displaystyle \frac{2}{5}\)
(e) None of these
Solution: (e)
Number of ways of selecting 2 red marbles from 6 red marbles = \(\displaystyle ^{6}{{C}_{2}}=15\)
Number of ways of selecting 2 marbles from urn = \(\displaystyle ^{{15}}{{C}_{2}}=105\)
Required Probability = \(\displaystyle \frac{{15}}{{105}}=\frac{1}{7}\)
3. If four marbles are picked at random, what is the probability that one is green, two are blue and one is red?
(a) \(\displaystyle \frac{3}{31}\)
(b) \(\displaystyle \frac{1}{5}\)
(c) \(\displaystyle \frac{18}{455}\)
(d) \(\displaystyle \frac{7}{15}\)
(e) None of these
Solution: (c)
Number of ways of selecting 2 blue and one yellow marble = \(\displaystyle ^{4}{{C}_{2}}{{\times }^{3}}{{C}_{1}}=6\times 3=18\)
Number of ways of selecting 3 marble from urn = \(\displaystyle ^{{15}}{{C}_{3}}=455\) Required Probability = \(\displaystyle \frac{{18}}{{455}}\)
4. If four marbles are picked at random, what is the probability that one is green, two are blue and one is red?
(a) \(\displaystyle \frac{24}{455}\)
(b) \(\displaystyle \frac{13}{35}\)
(c) \(\displaystyle \frac{11}{15}\)
(d) \(\displaystyle \frac{1}{3}\)
(e) None of these
Solution: (a)
Number of ways of selecting one green, two blue and one red marble = \(\displaystyle ^{2}{{C}_{1}}{{\times }^{4}}{{C}_{2}}{{\times }^{6}}{{C}_{1}}=2\times 6\times 6=72\)
Number of ways of selecting 4 marbles from urn = \(\displaystyle ^{{15}}{{C}_{4}}=\frac{{12\times 13\times 14\times 15}}{{4\times 3\times 2\times 1}}=1365\)
Required Probability = \(\displaystyle \frac{{72}}{{1365}}=\frac{{24}}{{455}}\)
5. If two marbles are picked at random, what is the probability that either both are green or both are yellow?
(a) \(\displaystyle \frac{5}{{91}}\)
(b) \(\displaystyle \frac{1}{{35}}\)
(c) \(\displaystyle \frac{1}{{3}}\)
(d) \(\displaystyle \frac{4}{{105}}\)
(e) None of these
Solution: (d)
Number of ways of selecting either two green marbles or two yellow marbles = \(\displaystyle ^{2}{{C}_{2}}{{+}^{3}}{{C}_{2}}=1+3=4\)
Number of ways of selecting 2 marbles = \(\displaystyle ^{{15}}{{C}_{2}}=105\)
Required Probability = \(\displaystyle \frac{4}{{105}}\)
