Class 10 Probability Word Problems — 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 bag has 5 red and 3 green marbles. Anil draws one marble at random. What is the probability that it is red?

A $\frac{5}{8}$
B $\frac{3}{8}$
C $\frac{5}{3}$
D $\frac{1}{5}$

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Answer: A)

\frac{5}{8}

Total = 5 + 3 = 8 marbles. P(red) =

\frac{5}{8}

2. A die is thrown once. What is the probability of getting a number that is neither prime nor composite?

A $\frac{1}{6}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{1}{4}$

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Answer: A)

\frac{1}{6}

Only 1 is neither prime nor composite. P =

\frac{1}{6}

3. A box has 4 black, 5 red and 6 white pens, total 15. A pen is drawn, its colour noted, and it is REPLACED. Then a second pen is drawn. What is the probability that both pens are red?

A $\frac{1}{9}$
B $\frac{2}{21}$
C $\frac{1}{3}$
D $\frac{25}{225}$

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Answer: A)

\frac{1}{9}

With replacement, draws are independent. P(red) =

\frac{5}{15} = \frac{1}{3}

each. P(both red) =

\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

4. Ishaan plays a board game using two dice of different colours, one yellow and one white. He throws both at the same time. What is the probability that the yellow die shows exactly twice as many dots as the white die?

A $\frac{1}{6}$
B $\frac{1}{18}$
C $\frac{1}{12}$
D $\frac{1}{36}$

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Answer: C)

\frac{1}{12}

Writing each throw as (white, yellow), the favourable throws are (1,2), (2,4) and (3,6); a white 4, 5 or 6 would need a yellow 8, 10 or 12, which no die has. So P =

\frac{3}{36}

\frac{1}{12}

\frac{1}{6}

also counts (2,1), (4,2) and (6,3), but in those it is the WHITE die that is twice the yellow;

\frac{1}{18}

misses one of the three throws;

\frac{1}{36}

names only a single throw.

5. A collection tin holds only one-rupee coins and two-rupee coins, worth 56 rupees altogether. There are 20 one-rupee coins in the tin. One coin is taken out at random. What is the probability that it is a two-rupee coin?

A $\frac{10}{19}$
B $\frac{1}{2}$
C $\frac{9}{19}$
D $\frac{9}{28}$

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Answer: C)

\frac{9}{19}

The 20 one-rupee coins are worth 20 rupees, so the two-rupee coins are worth 56 - 20 = 36 rupees, i.e. 36 / 2 = 18 coins. The tin holds 20 + 18 = 38 coins, so P =

\frac{18}{38}

\frac{9}{19}

\frac{10}{19}

answers for a one-rupee coin;

\frac{9}{28}

\frac{18}{56}

divides the coin count by the amount of money instead of by the number of coins;

\frac{1}{2}

assumes equal numbers of each coin.

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