From a set of 17 cards, numbered 1, 2, …. 17, one card is drawn. What is the probability that the number is a multiple of 3 or 7?
Answer: B
The total number of possible outcomes is 17
The number of favourable outcomes is 3, 6, 7, 9, 12, 14 and 15
= 7
Thus, the required probability = number of favourable outcomes/ total number of outcomes
= 7/17.
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Three coins are tossed. What is the probability of getting at most two tails?
Answer: A
It is given that 3 coins are tossed simultaneously.
=>Total outcomes when one coin is tossed=2
=>Total outcomes when 3 coins are tossed= 2n = 2 X 2 X 2
=8 outcomes
Favourable outcomes: TTH, THT, HTT, THH, THT,HHT,HHH=7
Therefore, Probability= 7/8
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What is the probability of getting a sum 9 from two throws of a dice?
Answer: C
Two throws give 36 outcomes. Those summing to 9 are
\[ (3,6),\ (4,5),\ (5,4),\ (6,3) \]
\[ P = \frac{4}{36} = \frac{1}{9} \]
Order matters, so \( (3,6) \) and \( (6,3) \) count separately.
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Each of letters of the word PILOTS is on separate cards, face down on the table. If you pick a card at random, what is the probability that the letter will be a T or an L?
Answer: B
There are 6 outcomes out of which 2 are favourable (which are, getting T or L).
Probability of an event E, P (E) =number of favourable outcomes/ total number of outcomes
Required probability = 2/6 = 1/3.
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Three unbiased coins are tossed. What is the probability of getting at most two heads?
Answer: D
"At most two heads" means everything except three heads, so use the complement.
\[ P(\text{3 heads}) = \frac{1}{8} \]
\[ P(\text{at most 2}) = 1 - \frac{1}{8} = \frac{7}{8} \]
Counting zero, one and two heads directly gives \( 1 + 3 + 3 = 7 \) of the 8 outcomes - the same answer for more work.
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What are the total outcomes when we throw three coins?
Answer: C
Each coin lands in one of two ways, and the three coins are independent, so the outcomes multiply.
\[ 2 \times 2 \times 2 = 2^{3} = 8 \]
Listing them: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. In general \( n \) coins give \( 2^{n} \) outcomes.
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A class consists of 15 girls and 10 boys. Three students are to be randomly selected. Find the probability that one boy and two girls are picked.
Answer: C
Choose one boy from 10 and two girls from 15, out of all ways to choose 3 from 25.
\[ P = \frac{{}^{10}C_{1} \times {}^{15}C_{2}}{{}^{25}C_{3}} = \frac{10 \times 105}{2300} \]
\[ = \frac{1050}{2300} = \frac{21}{46} \]
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The following probabilities are given; choose the correct answer for that which is not possible.
Answer: C
Every probability lies between 0 and 1 inclusive.
\[ 0 \leq P(E) \leq 1 \]
\[ 0.15 \ \checkmark \qquad \frac{2}{7} \approx 0.286 \ \checkmark \qquad \frac{7}{5} = 1.4 \ \times \]
\( \tfrac{7}{5} \) exceeds 1, so it cannot be a probability - nothing can be more certain than certain.
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When two coins are tossed simultaneously, what are the chances of getting at least one tail?
Answer: A
Two coins give four outcomes: HH, HT, TH, TT. Three of them contain a tail.
\[ P = \frac{3}{4} \]
Using the complement is quicker: the only outcome with no tail is HH, so
\[ P = 1 - \frac{1}{4} = \frac{3}{4} \]
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A coin is tossed five times. What is the probability that there is at least one tail?
Answer: A
Five tosses give \( 2^{5} = 32 \) outcomes. Only one of them - all heads - has no tail, so use the complement.
\[ P(\text{no tail}) = \frac{1}{32} \]
\[ P(\text{at least one tail}) = 1 - \frac{1}{32} = \frac{31}{32} \]
"At least one" is almost always easier through the complement than by counting one, two, three, four and five tails separately.
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