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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{66}} = \frac{1}{{11}}$$
52.
In a simultaneous throw of two coins, the probability of getting at least one head is-
(A) $$\frac{1}{2}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{2}{3}$$
(D) $$\frac{3}{4}$$
Solution:
Here S = {HH, HT, TH, TT} Let E = event of getting at least one head = {HT, TH, HH} $$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$
53.
A box contains 4 red, 5 green and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green ?
(A) $$\frac{2}{5}$$
(B) $$\frac{3}{5}$$
(C) $$\frac{1}{5}$$
(D) $$\frac{7}{15}$$
Solution:
Total number of balls = (4 + 5 + 6) = 15 P (drawing a red ball or a green ball) = P (red) + P (green) $$\eqalign{ & = \left( {\frac{4}{{15}} + \frac{5}{{15}}} \right) \cr & = \frac{9}{{15}} \cr & = \frac{3}{5} \cr} $$
54.
In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
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 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$
57.
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?
(A) $$\frac{{3}}{{4}}$$
(B) $$\frac{{4}}{{7}}$$
(C) $$\frac{{1}}{{8}}$$
(D) $$\frac{{3}}{{7}}$$
Solution:
Total number of balls = (6 + 8) = 14 Number of white balls = 8 P (drawing a white ball) = $$\frac{{8}}{{14}}$$ = $$\frac{{4}}{{7}}$$
58.
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} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{6} = \frac{1}{3}$$
59.
A speaks truth in 75% cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident ?
(A) 5%
(B) 15%
(C) 35%
(D) 45%
Solution:
Let $${{{\text{E}}_1}}$$ = event that A speaks the truth And $${{{\text{E}}_2}}$$ = event that B speaks the truth Then, $$\eqalign{ & P\left( {{E_1}} \right) = \frac{{75}}{{100}} = \frac{3}{4}, \cr & P\left( {{E_2}} \right) = \frac{{80}}{{100}} = \frac{4}{5}, \cr & P\left( {{{\overline E }_1}} \right) = \left( {1 - \frac{3}{4}} \right) = \frac{1}{4}, \cr & P\left( {{{\overline E }_2}} \right) = \left( {1 - \frac{4}{5}} \right) = \frac{1}{5} \cr} $$ P (A and B contradict each other) = P [(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)] $$\eqalign{ & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right)or\left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left[ {\left( {{E_1} \cap {{\overline E }_2}} \right) + \left( {{{\overline E }_1} \cap {E_2}} \right)} \right] \cr & = P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right) + P\left( {{{\overline E }_1}} \right).P\left( {{E_2}} \right) \cr & = \left( {\frac{3}{4} \times \frac{1}{5}} \right) + \left( {\frac{1}{4} \times \frac{4}{5}} \right) \cr & = \left( {\frac{3}{{20}} + \frac{1}{5}} \right) \cr & = \frac{7}{{20}} \cr & = \left( {\frac{7}{{20}} \times 100} \right)\% \cr & = 35\% \cr} $$
60.
The probability that a card drawn from a pack of 52 cards will be a diamond or a king is -
(A) $$\frac{{2}}{{13}}$$
(B) $$\frac{{4}}{{13}}$$
(C) $$\frac{{1}}{{13}}$$
(D) $$\frac{{1}}{{52}}$$
Solution:
Here, n(S) = 52 There are 13 cards of diamond (including one king) and there are three more kings. Let E = event of getting a diamond or a king Then, n(E) = (13 + 3) = 16 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$