C

We have n(s) = [latex]^{52}{C}_{2} 52 = 52 \times \frac {51}{2 \times 1} = 1326[/latex]
Let A = event of getting both black cards
B = event of getting both queens
A∩B = event of getting queen of black cards
n(A) = [latex]\frac {52 \times 51}{2 \times 1} = ^{26}{C}_{2} = 325, n(B)= \frac{26 \times 25 }{2 \times 1}= \frac {4 \times 3}{2 \times 1} = 6 and n(A∩B) = ^{4}{C}_{2} = 1 [/latex]
P(A) = [latex]\frac {n(A)}{n(S)} = \frac {325}{1326}[/latex]
P(B) = [latex]\frac {n(B)}{n(S)} = \frac {6}{1326}[/latex]
P(A∩B) = [latex]\frac {n(A∩B)}{n(S)} = \frac {1}{1326}[/latex]
P(A∪B) = P(A) + P(B) - P(A∩B) = [latex]\frac {(325 + 6 - 1)}{1326} = \frac {330}{1326} = \frac {55}{221}[/latex]

A

Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = [latex]\frac {n(E)}{n(S)} = \frac {15}{36} = \frac {5}{12}[/latex]

A

Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15 = [latex] ^{15}{C}_{3} = \frac {15 \times 14 \times 13}{3 \times 2 \times 1} = 455[/latex]
Let E = event of getting all the 3 red balls.
n(E) = [latex] ^{5}{C}_{3} = \frac {5 \times 4 }{3 \times 2 \times 1} = 10 [/latex]
=> P(E) = [latex]\frac {n(E)}{n(S)} \frac {10}{455} = \frac {2}{91}[/latex]

A

Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
P(E) = [latex]\frac {n(E)}{n(s)} = \frac {10}{35} = \frac {2}{7}[/latex]

A

Let, S - sample space E - event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= [latex]^{25}{C}_{3}[/latex]
= 2300.
n(E) = [latex]^{10}{C}_{1}[/latex] × [latex]^{15}{C}_{2}[/latex] = 1050.
P(E) = [latex]\frac {n(E)}{n(s)} = \frac {1050}{2300} = \frac {21}{46}[/latex]