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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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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A fair dice is rolled. Probability of getting a number x such that 1 ≤ x ≤ 6, is
Answer: C
Every face of a die shows a number \( x \) with \( 1 \leq x \leq 6 \), so the event covers all six outcomes.
\[ P = \frac{6}{6} = 1 \]
An event that is bound to happen is a sure event, with probability exactly 1. No probability can exceed 1.
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A letter is chosen at random from the letters of the word ASSASSINATION. The probability that the letter chosen is a consonant is:
Answer: B
ASSASSINATION has 13 letters: A, S, S, A, S, S, I, N, A, T, I, O, N.
\[ \text{vowels: A, A, A, I, I, O} \rightarrow 6 \]
\[ \text{consonants: S, S, S, S, N, N, T} \rightarrow 7 \]
\[ P(\text{consonant}) = \frac{7}{13} \]
The two counts must add to 13, which is the check that neither has been miscounted.
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In a Poisson probability distribution
Answer: A
For a Poisson distribution with parameter \( \lambda \),
\[ \text{mean} = \lambda, \qquad \text{variance} = \lambda \]
so the two are always equal - a property unique to the Poisson among the standard discrete distributions, and the usual quick test of whether count data are Poisson.
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A bucket contains 10 brown balls, 8 green balls, and 12 red balls and you pick one randomly without looking. What is the probability that the ball will be brown?
Answer: C
The bucket holds \( 10 + 8 + 12 = 30 \) balls.
\[ P = \frac{10}{30} = \frac{1}{3} \]
The decimal 0.33 is only an approximation of \( \tfrac{1}{3} \), which is the exact value.
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The joint probability is
Answer: A
Joint probability is the probability that two events occur together, written \( P(A \cap B) \).
The second option describes conditional probability, \( P(A \mid B) \); mutual exclusivity would make the joint probability zero; and prior probability is the belief held before new evidence, a term from Bayes' theorem.
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If P (A) and P (not A) are complementary events and P (A) = 0.15, then P (not A) =?
Answer: C
Complementary events cover every possibility between them, so their probabilities add to 1.
\[ P(\text{not } A) = 1 - P(A) = 1 - 0.15 = 0.85 \]
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Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
Northern Coalfields Limited - 2020, NCL Technician - 2020,
Answer: D
Count each set, then remove the overlap.
\[ \text{multiples of 3} = 6, \qquad \text{multiples of 5} = 4, \qquad \text{multiples of 15} = 1 \]
\[ 6 + 4 - 1 = 9 \quad\Longrightarrow\quad P = \frac{9}{20} \]
15 is a multiple of both and would otherwise be counted twice.
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