A die is thrown once, the probability of getting a composite number on the die is
Answer: A
Of the faces 1 to 6, the composite numbers are 4 and 6.
\[ P = \frac{2}{6} = \frac{1}{3} \]
2, 3 and 5 are prime, and 1 is neither prime nor composite - including it would give \( \tfrac{1}{2} \).
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If you roll a pair of dice, what is the probability that (at least) one of the dice is a 4 or the sum of the dice is 7?
Answer: D
Let \( A \) be "at least one die shows 4" and \( B \) be "the sum is 7", over the 36 outcomes.
\[ |A| = 11 \quad (\text{six with the first a 4, six with the second, less the double-counted } (4,4)) \]
\[ |B| = 6: \ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \]
\[ |A \cap B| = 2: \ (3,4),(4,3) \]
\[ |A \cup B| = 11 + 6 - 2 = 15 \quad\Longrightarrow\quad P = \frac{15}{36} \]
Both corrections matter: \( (4,4) \) within \( A \), and the two outcomes shared by \( A \) and \( B \).
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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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What is the probability of getting all heads or all tails, when three coins are tossed simultaneously ?
Answer: C
When three coins are tossed simultaneously, there are 8 possible outcomes, which are (HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT), where H is represents the head and T represents the tail.
Favorable outcomes of getting all heads or all tails are HHH and TTT.
We know that, Probability of an event E, P (E) =Number of favorable outcomes /Total number of outcomes
Hence, the required probability = 2/8 = 1/4
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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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In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
Answer: A
"Neither red nor green" leaves only the blue balls.
\[ \text{total} = 8 + 7 + 6 = 21 \]
\[ P = \frac{7}{21} = \frac{1}{3} \]
Recognising the complement as a single colour turns a two-case count into one division.
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In the special rule of addition of probability, the events are always
Answer: B
The special rule of addition is
\[ P(A \cup B) = P(A) + P(B) \]
and it holds only when the events are mutually exclusive - they cannot both occur, so \( P(A \cap B) = 0 \) and there is no overlap to subtract.
Independence is the condition for the special rule of multiplication. The general addition rule, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \), needs no condition at all.
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What is the possibility of having 53 Thursdays in a non-leap year?
Answer: B
A non-leap year has 365 days:
\[ 365 = 52 \times 7 + 1 \]
So there are 52 Thursdays for certain, plus one leftover day. A 53rd Thursday occurs only if that day is a Thursday, and it is equally likely to be any of the seven days.
\[ P = \frac{1}{7} \]
A leap year has two leftover days and the probability rises to \( \tfrac{2}{7} \).
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The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow
Answer: C
An event and its complement together account for every possibility, so their probabilities add to 1.
\[ P(\text{not rain}) = 1 - P(\text{rain}) = 1 - 0.85 = 0.15 \]
\[ 0.15 = \frac{15}{100} = \frac{3}{20} \]
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A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Answer: A
The bag holds \( 2 + 3 + 2 = 7 \) balls, of which 5 are not blue.
\[ P = \frac{{}^{5}C_{2}}{{}^{7}C_{2}} = \frac{10}{21} \]
Counting the non-blue balls directly is quicker than working through the red and green cases separately.
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