Class 10 Probability (Theoretical) — practice questions with answers

Statistics · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. A coin is tossed. What is the probability of getting heads?

A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D1

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Answer: A)

\frac{1}{2}

A coin has two equally likely sides, so the probability of getting heads is

\frac{1}{2}

2. A card is drawn from a well-shuffled deck of 52 cards. What is the probability of getting an ace?

A $\frac{1}{13}$
B $\frac{1}{4}$
C $\frac{1}{52}$
D $\frac{4}{13}$

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Answer: A)

\frac{1}{13}

There are 4 aces in a deck of 52 cards. Probability =

\frac{4}{52} = \frac{1}{13}

3. A bag contains 3 red, 5 black, and 7 white balls. A ball is drawn at random. What is the probability that the ball is red or white?

A $\frac{2}{3}$
B $\frac{1}{3}$
C $\frac{7}{15}$
D $\frac{8}{15}$

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Answer: A)

\frac{2}{3}

Total = 3 + 5 + 7 = 15. Red or white = 3 + 7 = 10. Probability =

\frac{10}{15} = \frac{2}{3}

4. One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that it is neither an ace nor a king?

A $\frac{2}{13}$
B $\frac{1}{13}$
C $\frac{12}{13}$
D $\frac{11}{13}$

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Answer: D)

\frac{11}{13}

There are 4 aces and 4 kings, so 8 cards must be excluded. Favourable cards = 52 - 8 = 44. Probability =

\frac{44}{52}

\frac{11}{13}

. The value

\frac{2}{13}

is the probability of drawing an ace or a king, which is the complement.

5. A survey report states that the probability that a student likes cricket is 0.65 and that the probability that the same student does not like cricket is 0.45. Is this report possible?

ANo, because an event and its complement must have probabilities adding up to 1, but 0.65 + 0.45 = 1.1
BYes, because both values lie between 0 and 1, so both are valid probabilities
CYes, because a student may like cricket on some days and not on others
DNo, because the probability of liking cricket must always be 0.5

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Answer: A) No, because an event and its complement must have probabilities adding up to 1, but 0.65 + 0.45 = 1.1

Liking cricket and not liking cricket are complementary events, so P(E) + P(not E) must equal 1. Here the two reported values add to 1.1, which is impossible, so the report cannot be correct.

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