In a simultaneous throw of two dice, what is the probability of getting a total of 7?
(A)
(B)
(C)
(D)
Solution:
We know that in a simultaneous throw of two dice, n(S) = 6 × 6 = 36 Let E = event of getting a total of 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{36}} = \frac{1}{6}$$
22.
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 $$\eqalign{ & = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} \cr & = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} \cr & = \frac{4}{6} \cr & = \frac{2}{3} \cr} $$
23.
In a simultaneous throw of two dice, what is the probability of getting a doublet ?
(A)
(B)
(C)
(D)
Solution:
In a simultaneous throw of dice, n (S) = (6 × 6) = 36 Let E = event of getting a doublet = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] ∴ P(E) = $$\frac{{n (E)}}{{n (S)}}$$ = $$\frac{{6}}{{36}}$$ = $$\frac{{1}}{{6}}$$
24.
A man and his wife appear in an interview for two vacancies in the same post. The probability two of husbandÃÂÃÂÃÂâÃÂÃÂÃÂÃÂs selection is and the probability of wifeÃÂÃÂÃÂâÃÂÃÂÃÂÃÂs selection is. What is the probability that only one of them is selected?
(A)
(B)
(C)
(D)
Solution:
Let $${E_1}$$ = event that the husband is selected and $${E_2}$$ = event that the wife is selected. Then, $$P({E_1}) = \frac{1}{7}{\text{ }}$$ and $${\text{ }}P({E_2}) = \frac{1}{5}$$ ∴ $$P({\overline E _1}) = \left( {1 - \frac{1}{7}} \right)$$ $${\text{ = }}\frac{6}{7}$$ and $$P({\overline E _2}) = $$ $$\left( {1 - \frac{1}{5}} \right) = \frac{4}{5}$$ ∴ Required probability = P [( A and not B) or (B and not A)] = P $$\left[ {\left( {{E_1} \cap {{\bar E}_2}} \right)\,{\text{or}}\,\left( {{E_2} \cap {{\bar E}_1}} \right)} \right]$$ = P $$\left( {{E_1} \cap {{\overline E }_2}} \right) + {\text{ P}}\left( {{E_2} \cap {{\overline E }_1}} \right)$$= $${\text{P}}\left( {{E_1}} \right).{\text{ P}}\left( {{{\overline E }_2}} \right) + P\left( {{E_2}} \right).P\left( {{{\overline E }_1}} \right)$$ $$\eqalign{ & = \left( {\frac{1}{7} \times \frac{4}{5}} \right) + \left( {\frac{1}{5} \times \frac{6}{7}} \right) \cr & = \frac{10}{{35}} \cr & = \frac{2}{7} \cr} $$
25.
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)
(B)
(C)
(D)
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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$
26.
A speaks truth in 60% cases B speaks truth in 70% cases. The probability that they will way say the same thing while describing a single event, is-
(A) 0.54
(B) 0.56
(C) 0.68
(D) 0.94
Solution:
Let $${E_1}$$ = event that A speaks the truth And $${E_2}$$ = event that B speaks the truth Then, $$\eqalign{ & P\left( {{E_1}} \right) = \frac{{60}}{{100}} = \frac{3}{5}, \cr & P\left( {{E_2}} \right) = \frac{{70}}{{100}} = \frac{7}{{10}}, \cr & P\left( {{{\bar E}_1}} \right) = \left( {1 - \frac{3}{5}} \right) = \frac{2}{5}, \cr & P\left( {{{\bar E}_2}} \right) = \left( {1 - \frac{7}{{10}}} \right) = \frac{3}{{10}} \cr} $$ P (A and B say the same thing) = P [(A speaks the truth and B speaks the truth) or (A tells a lie and B tells a lie)] $$=$$ P [$$\left( {{E_1} \cap {E_2}} \right)$$ or $$({\overline E _1} \cap {\overline E _2})] $$ $$=$$ P $$\left( {{E_1} \cap {E_2}} \right)$$ + $$({\overline E _1} \cap {\overline E _2})$$ $$=$$ P ($${{E_1}}$$). P ($${{E_2}}$$) + P ($${\overline E _1}$$) . P ($${\overline E _2}$$) $$\eqalign{ & = \left( {\frac{3}{5} \times \frac{7}{{10}}} \right) + \left( {\frac{2}{5} \times \frac{3}{{10}}} \right) \cr & = \frac{{27}}{{50}} \cr & = 0.54 \cr} $$
27.
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3 ?
(A)
(B)
(C)
(D)
Solution:
Here, S = {1, 2, 3, 4,........, 19, 20} Let E = even of getting a multiple of 3 = {3, 6, 9, 12, 15, 18} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{20}} = \frac{3}{{10}}$$
28.
A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If three marbles are picked up at random, what is the probability that 2 are blue and 1 is yellow ?
(A)
(B)
(C)
(D)
Solution:
Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 blue and 1 yellow marble. Then, n(E) = $$\left( {{}^4\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right)$$ $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times 3} \right)$$ = 18 Also, n(S) = $${}^{15}\mathop C\nolimits_3 = $$ $$\frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$ = 455 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{18}}{{455}}$$
30.
A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?