C

From 1 to 24, there are 8 numbers which are multiple of 3
Case 1: 2 are multiple of 3, and one any other number from (24 - 8) = 16 tickets
[latex]\frac {^{8}{C}_{2} \times ^{16}{C}_{1}}{^{24}{C}_{3}} = \frac {56}{253}[/latex]
Case 2: all are multiples of 3.
[latex]\frac {^{8}{C}_{3}}{^{24}{C}_{3}} = \frac {7}{253}[/latex]
Add both cases.
[latex] \frac {56}{253} + \frac {7}{253} = \frac {63}{253}[/latex]

B

Total balls = 15
Not red means green or blue i.e. any of (5 + 6) = 11 balls
So prob. = [latex]\frac {^{11}{C}_{2}}{^{15}{C}_{2}} = \frac {11}{21}[/latex]

A

Prob. of black card:
[latex]\frac {^{26}{C}_{2}}{^{52}{C}_{2}} = \frac {25}{102}[/latex]
Prob. of heart card:
[latex]\frac {^{13}{C}_{2}}{^{52}{C}_{2}} = \frac {3}{51}[/latex]
Add both cases.
[latex]\frac {25}{102} + \frac {3}{51} = \frac {31}{102}[/latex]

D

There are total 26 cards black, and 4 jacks in which 2 are black jacks
So case 1: both are black [latex]\frac {^{26}{C}_{2}}{^{52}{C}_{2}}[/latex]
case 2: both are jack [latex]\frac {^{4}{C}_{2}}{^{52}{C}_{2}}[/latex]
Add both cases.
But now 2 black jacks have been added in both cases, so subtracting their prob. : [latex]\frac {^{2}{C}_{2}}{^{52}{C}_{2}}[/latex]
So [latex]\frac {325}{1326} + \frac {6}{1326} – \frac {1}{1326} = \frac {55}{221}[/latex]

B

Case 1: Prob. when 2 men, 1 woman and 1 child [latex]\frac {^{3}{C}_{2} \times ^{4}{C}_{1} \times ^{2}{C}_{1}}{^{9}{C}_{4}} = \frac {4}{21}[/latex]
Case 2: Prob. when 1 man, 2 women and 1 child [latex]\frac {^{3}{C}_{1} \times ^{4}{C}_{2} \times ^{2}{C}_{1}}{^{9}{C}_{4}} = \frac {2}{7}[/latex]
Case 3: Prob. when 1 man, 1 woman and 2 children [latex]\frac {^{3}{C}_{1} \times ^{4}{C}_{1} \times ^{2}{C}_{2}}{^{9}{C}_{4}} = \frac {2}{21}[/latex]
Add three cases
[latex]\frac {4}{21} + \frac {2}{7} + \frac {2}{21} = \frac {12}{21}[/latex]