A box has 6 black, 4 red, 2 white and 3 blue shirts. Find the probability of drawing 2 black shirts if they are picked randomly?
Answer: D
\[ P = \frac{{}^{6}C_{2}}{{}^{15}C_{2}} = \frac{15}{105} = \frac{1}{7} \]
Equivalently, drawing one after the other: \( \tfrac{6}{15} \times \tfrac{5}{14} = \tfrac{30}{210} = \tfrac{1}{7} \). The \( \tfrac{6}{15} \) option ignores the second draw entirely.
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In a set of 30 game cards, 17 are white and rest are green. 4 white and 5 green are marked IMPORTANT. If a card is chosen randomly from this set, what is the possibility of choosing a green card or an ‘IMPORTANT’ card?
Answer: C
Of the 30 cards, 17 are white and 13 green. Nine are marked IMPORTANT: 4 white and 5 green.
Adding the two counts directly would double-count the 5 green IMPORTANT cards, so use the addition rule:
\[ P(G \cup I) = P(G) + P(I) - P(G \cap I) \]
\[ = \frac{13}{30} + \frac{9}{30} - \frac{5}{30} = \frac{17}{30} \]
Equivalently, count the 13 green cards plus the 4 white IMPORTANT ones: \( \tfrac{17}{30} \).
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There are 2 pots. One pot has 5 red and 3 green marbles. Other has 4 red and 2 green marbles. What is the probability of drawing a red marble?
Answer: B
A pot is chosen first, each with probability \( \tfrac{1}{2} \), and then a marble from it.
\[ P = \frac{1}{2} \times \frac{5}{8} + \frac{1}{2} \times \frac{4}{6} \]
\[ = \frac{5}{16} + \frac{1}{3} = \frac{15}{48} + \frac{16}{48} = \frac{31}{48} \]
Pooling all 14 marbles into one pot would give \( \tfrac{9}{14} \) - the first option - but that would treat the two pots as equally stocked, which they are not.
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A pot has 2 white, 6 black, 4 grey and 8 green balls. If one ball is picked randomly from the pot, what is the probability of it being black or green?
Answer: B
The pot holds \( 2 + 6 + 4 + 8 = 20 \) balls.
Black and green are mutually exclusive, so their probabilities add:
\[ P = \frac{6}{20} + \frac{8}{20} = \frac{14}{20} = \frac{7}{10} \]
No ball is both black and green, so no overlap has to be subtracted.
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A box has 6 black, 4 red, 2 white and 3 blue shirts. When 2 shirts are picked randomly, what is the probability that either both are white or both are blue?
Answer: A
The two cases cannot both happen, so add their counts.
\[ \text{both white} = {}^{2}C_{2} = 1, \qquad \text{both blue} = {}^{3}C_{2} = 3 \]
\[ P = \frac{1 + 3}{{}^{15}C_{2}} = \frac{4}{105} \]
With only two white shirts there is exactly one way to take both, which is what keeps the numerator so small.
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What will be the possibility of drawing a jack or a spade from a well shuffled standard deck of 52 playing cards?
Answer: A
There are 4 jacks and 13 spades, but the jack of spades is in both groups and must not be counted twice.
\[ P(J \cup S) = P(J) + P(S) - P(J \cap S) \]
\[ = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]
Adding without the correction gives \( \tfrac{17}{52} \) - the last option.
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Suresh keeps all his socks in a single drawer. He has 24 pairs of white socks and 18 pairs of grey socks. Suresh picks 3 socks randomly. Find the possibility of Suresh choosing a matching pair.
Answer: D
There are only two colours of sock in the drawer, so consider the worst case.
The first two socks might be one white and one grey. But the third must match one of them, since there is no third colour available.
\[ P = 1 \]
This is the pigeonhole principle: three items sorted into two categories force at least two into the same one. The actual numbers of pairs never enter the argument.
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Three unbiased coins are tossed. What is the probability of getting at least 2 tails?
Answer: B
Three coins give \( 2^{3} = 8 \) equally likely outcomes. Those with at least two tails are
\[ \text{TTH},\ \text{THT},\ \text{HTT},\ \text{TTT} \]
\[ P = \frac{4}{8} = 0.5 \]
"At least two" takes in all three tails as well; counting only exactly two would give 0.375.
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Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?
Answer: A
The two selections are independent, so the probabilities multiply.
\[ P = \frac{2}{5} \times \frac{4}{7} = \frac{8}{35} \]
Adding them would answer a different question - and would exceed 1 in many such problems, which is the quickest sign that multiplication is meant.
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Tickets numbered 1 to 50 are mixed and one ticket is drawn at random. Find the probability that the ticket drawn has a number which is a multiple of 4 or 7?
Answer: A
Count each set, then remove the overlap.
\[ \text{multiples of 4} = \left\lfloor \tfrac{50}{4} \right\rfloor = 12, \qquad \text{multiples of 7} = \left\lfloor \tfrac{50}{7} \right\rfloor = 7 \]
\[ \text{multiples of 28} = \left\lfloor \tfrac{50}{28} \right\rfloor = 1 \]
\[ 12 + 7 - 1 = 18 \quad\Longrightarrow\quad P = \frac{18}{50} = \frac{9}{25} \]
28 is the only number counted twice, being a multiple of both 4 and 7.
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