In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
(A) $$\frac{{1}}{{3}}$$
(B) $$\frac{{3}}{{4}}$$
(C) $$\frac{{7}}{{19}}$$
(D) $$\frac{{8}}{{21}}$$
Solution:
Total number of balls = (8 + 7 + 6) = 21 Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue $$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$
42.
A bag contains 3 blue, 2 green and 5 red balls. If four balls are picked at random, what is the probability that two are green and two are blue?
(A) $$\frac{1}{18}$$
(B) $$\frac{1}{70}$$
(C) $$\frac{3}{5}$$
(D) $$\frac{1}{2}$$
Solution:
Number of blue balls = 3 Number of green balls = 2 Numbers of red balls = 5 Total balls in the bag = 3 + 2 + 5 = 10 Total possible outcomes = Selection of 4 balls out of 10 balls $$ = {}^{10}\mathop C\nolimits_4 = \frac{{10!}}{{4! \times (10 - 4)!}}$$ $$ = \frac{{10 \times 9 \times 8 \times 7}}{{1 \times 2 \times 3 \times 4}}$$ = 210 Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls) $$\eqalign{ & = {}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_2 \cr & = 1 \times 3 \cr & = 3 \cr} $$ ∴ Required probability $$ = \frac{{{\text{Favorable outcomes}}}}{{{\text{Total possible outcomes}}}}$$ $$\eqalign{ & = \frac{3}{{210}} \cr & = \frac{1}{{70}} \cr} $$
43.
Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are king, is-
(A) $$\frac{7}{13}$$
(B) $$\frac{3}{26}$$
(C) $$\frac{63}{221}$$
(D) $$\frac{55}{221}$$
Solution:
Clearly, n (S) = $$n{\text{ }}(S) = $$ $${}^{52}\mathop C\nolimits_2 = $$ $$\frac{{\left( {52 \times 51} \right)}}{2}$$ = 1326 Let $${{E_1}}$$ = event of getting both red cards $${{E_2}}$$ = event of getting both kings Then, $${{E_1}}$$ $$ \cap $$ $${{E_2}}$$ = event of getting 2 kings of red cards. ∴ $$n{\text{ }}({E_1}) = {}^{26}\mathop C\nolimits_2 = \frac{{\left( {26 \times 25} \right)}}{{\left( {2 \times 1} \right)}}$$ = 325 and $$n{\text{ }}({E_2}) = {}^4\mathop C\nolimits_2 = \frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$ = 6 $$n\left( {{E_1} \cap {E_2}} \right) = {}^2{C_2} = 1$$ $$\therefore P({E_1}) = \frac{{n({E_1})}}{{n(S)}} = \frac{{325}}{{1326}}$$ and $$P({E_2}) = \frac{{n({E_2})}}{{n(S)}} = \frac{6}{{1326}}$$ $$P({E_1} \cap {E_2}) = \frac{1}{{1326}}$$ ∴ P (both red or both kings) $$\eqalign{ & = P\left( {{E_1} \cup {E_2}} \right) \cr & = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr & = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr & = \frac{{330}}{{1326}} \cr & = \frac{{55}}{{221}} \cr} $$
44.
A basket contains 4 red, 5 blue and 3 green marbles. If 2 marbles are drawn at random from the basket, What is the probability that both are red ?
(A) $$\frac{{3}}{{7}}$$
(B) $$\frac{{1}}{{2}}$$
(C) $$\frac{{1}}{{11}}$$
(D) $$\frac{{1}}{{6}}$$
Solution:
Total number of balls = (4 + 5 + 3) = 12 Let E be the event of drawing 2 red balls. Then, n (E) = $${}^4\mathop C\nolimits_2 = \frac{{4 \times 3}}{{2 \times 1}}$$ = 6 Also n (S) = $${}^{12}\mathop C\nolimits_2 = \frac{{12 \times 11}}{{2 \times 1}}$$ = 66 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{66}} = \frac{1}{{11}}$$
45.
A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least 1 woman ?
(A) $$\frac{{1}}{{10}}$$
(B) $$\frac{{9}}{{20}}$$
(C) $$\frac{{1}}{{20}}$$
(D) $$\frac{{9}}{{10}}$$
Solution:
Total number of persons = (3 + 2) = 5 $$\therefore n(S) = {}^5\mathop C\nolimits_3 = {}^5\mathop C\nolimits_2 $$ $$ = \frac{{5 \times 4}}{{2 \times 1}}$$ = 10 Let E be the event of selecting 3 members having at least 1 women Then, n(E) = n [(1 women and 2 men ) or (2 women and 1 man)] = n (1 woman and 2 men) + n (2 women and 1 man) $$\eqalign{ & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_2 } \right) + \left( {{}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_1 } \right) + \left( {1 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {2 \times 3} \right) + \left( {1 \times 3} \right) \cr & = \left( {6 + 3} \right) \cr & = 9 \cr} $$ $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{10}}$$
46.
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
(A) $$\frac{{1}}{{15}}$$
(B) $$\frac{{25}}{{57}}$$
(C) $$\frac{{35}}{{256}}$$
(D) $$\frac{{1}}{{221}}$$
Solution:
Let S be the sample space $$\eqalign{ & {\text{Then}},n\left( S \right) = {}^{52}{C_2} \cr & = \frac{{ {52 \times 51} }}{{\left( {2 \times 1} \right)}} \cr & = 1326 \cr} $$ Let E = event of getting 2 kings out of 4 $$\eqalign{ & \therefore n\left( E \right) = {}^4{C_2} = \frac{{ {4 \times 3} }}{{ {2 \times 1} }} = 6 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{6}{{1326}} \cr & = \frac{1}{{221}} \cr} $$
47.
Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is-
(A) $$\frac{{1}}{{9}}$$
(B) $$\frac{{1}}{{5}}$$
(C) $$\frac{{1}}{{12}}$$
(D) $$\frac{{10}}{{21}}$$
Solution:
n(S) = number of ways of choosing 4 persons out of 9 $$ = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126 n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal n(E) $$ = \left( {{}^4\mathop C\nolimits_2 \times {}^5\mathop C\nolimits_2 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$ = 60 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$
48.
In a single throw of die, what is the probability of getting a number greater than 4 ?
(A) $$\frac{1}{2}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{2}{3}$$
(D) $$\frac{1}{4}$$
Solution:
When a die is thrown, we have S = {1, 2, 3, 4, 5, 6} Let, E = event of getting a number greater than 4 = {5, 6} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{6} = \frac{1}{3}$$
49.
A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
(A) $$\frac{{1}}{{22}}$$
(B) $$\frac{{3}}{{22}}$$
(C) $$\frac{{2}}{{91}}$$
(D) $$\frac{{2}}{{77}}$$
Solution:
Let S be the sample space Then, n(S) = number of ways of drawing 3 balls out of 15 $$\eqalign{ & = {}^{15}{C_3} \cr & = \frac{{ {15 \times 14 \times 13} }}{{ {3 \times 2 \times 1} }} \cr & = 455 \cr} $$ Let E = event of getting all the 3 red balls $$\eqalign{ & \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr & = \frac{{ {5 \times 4} }}{{ {2 \times 1} }} = 10 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{10}}{{455}} \cr & = \frac{2}{{91}} \cr} $$
50.
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow ?
(A) $$\frac{5}{91}$$
(B) $$\frac{1}{35}$$
(C) $$\frac{1}{3}$$
(D) $$\frac{4}{105}$$
Solution:
Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 marbles such that either both are green or both are yellow. Then, n (E) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$ = (1 + 3) = 4And, n (S) = $${}^{15}\mathop C\nolimits_2 = $$ $$\frac{{15 \times 14}}{{2 \times 1}}$$ = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$