An urn contains 6 red, 4 blue, 2 green 3 yellow marbles. If two marbles are drawn at random from the run, what is the probability that both are red ?
(A)
(B)
(C)
(D)
Solution:
Total number of balls = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 red balls. Then, n(E) $$ = {}^6\mathop C\nolimits_2 $$ $$ = \frac{{6 \times 5}}{{2 \times 1}}$$ = 15 Also, $$n(S) = {}^{15}\mathop C\nolimits_2 $$ $$ = \frac{{15 \times 14}}{{2 \times 1}}$$ = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{105}} = \frac{1}{7}$$
2.
Dev can hit a target 3 times in 6 shots pawan can hit the target 2 times in 6 shots and Lakhan can hit the target 4 times in 4 shots. What is the probability that at least 2 shots hit the target -
(A)
(B)
(C)
(D) None of these
Solution:
Probability of hitting the target: Dev can hit target ⇒ $$\frac{{3}}{{6}}$$ =$$\frac{{1}}{{2}}$$ Lakhan can hit target =$$\frac{{4}}{{4}}$$ = 1 Pawan can hit target = $$\frac{{2}}{{6}}$$ = $$\frac{{1}}{{3}}$$ Required probability that at least 2 shorts hit target $$\eqalign{ & = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} \cr & = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} \cr & = \frac{4}{6} \cr & = \frac{2}{3} \cr} $$
3.
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card ?
(A)
(B)
(C)
(D)
Solution:
Clearly, there are 52 cards, out of which there are 12 face cards 4 jack, 4 queens, and 4 kings ∴ P (getting a face card) $$ = \frac{{12}}{{52}}$$ $$ = \frac{3}{{13}}$$
4.
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)
(B)
(C)
(D)
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) $$\eqalign{ & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_2 } \right) + \left( {{}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {{}^2\mathop C\nolimits_1 \times {}^3\mathop C\nolimits_1 } \right) + \left( {1 \times {}^3\mathop C\nolimits_1 } \right) \cr & = \left( {2 \times 3} \right) + \left( {1 \times 3} \right) \cr & = \left( {6 + 3} \right) \cr & = 9 \cr} $$ $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{9}{{10}}$$
5.
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 blue nor green ?
(A)
(B)
(C)
(D)
Solution:
Total number of balls = (8 + 7 + 6) = 21 Let E = event that the ball drawn is neither blue nor green = Even that the ball drawn is red ∴ n(E) = 8 P(E) = $$\frac{{8}}{{21}}$$
6.
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
(A)
(B)
(C)
(D)
Solution:
Clearly, there are 52 cards, out of which there are 12 face cards. $$\eqalign{ & \therefore P\left( {{\text{getting}}\,{\text{a}}\,{\text{face}}\,{\text{card}}} \right) \cr & = \frac{{12}}{{52}} = \frac{3}{{13}} \cr} $$
7.
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card ?
(A)
(B)
(C)
(D)
Solution:
Clearly, there are 52 cards, out of which there are 12 face cards 4 jack, 4 queens, and 4 kings ∴ P (getting a face card) $$ = \frac{{12}}{{52}}$$ $$ = \frac{3}{{13}}$$
8.
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)
(B)
(C)
(D)
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 $$ = \frac{{{\text{Favorable outcomes}}}}{{{\text{Total possible outcomes}}}}$$ $$\eqalign{ & = \frac{3}{{210}} \cr & = \frac{1}{{70}} \cr} $$
9.
Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are king, is-
(A)
(B)
(C)
(D)
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) $$\eqalign{ & = P\left( {{E_1} \cup {E_2}} \right) \cr & = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr & = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr & = \frac{{330}}{{1326}} \cr & = \frac{{55}}{{221}} \cr} $$
10.
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)
(B)
(C)
(D)
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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{60}}{{126}} = \frac{{10}}{{21}}$$