Probability Practice Questions with Solutions for Class 10

True or false#

Ques­tion 1#

State true or false, with a rea­son: the prob­a­bil­ity of an event can be $\displaystyle \frac{6}{5}$.

The num­ber of favourable out­comes can never exceed the total num­ber of out­comes, so every prob­a­bil­ity is at most 1. Since $\displaystyle \frac{6}{5} \gt 1$, it can­not be a prob­a­bil­ity.

Answer: False.

Ques­tion 2#

State true or false, with a rea­son: when three coins are tossed, the prob­a­bil­ity of exactly two heads is $\displaystyle \frac{1}{4}$, because the num­ber of heads can only be 0, 1, 2 or 3.

Those four results are not equally likely. The 8 equally likely out­comes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT, and exactly two heads occur in HHT, HTH and THH, so the prob­a­bil­ity is $\displaystyle \frac{3}{8}$.

Answer: False; the prob­a­bil­ity is $\displaystyle \frac{3}{8}$.

Ques­tion 3#

State true or false: if the prob­a­bil­ity that it rains tomor­row is 0.36, the prob­a­bil­ity that it does not rain tomor­row is 0.64.

"Rain" and "no rain" are com­ple­men­tary, so $P(\text{no rain}) = 1 - 0.36 = 0.64$.

Answer: True.

Ques­tion 4#

State true or false: when two dice are rolled, get­ting a sum of 13 is an impos­si­ble event.

The largest pos­si­ble sum is $6 + 6 = 12$, so no out­come gives 13. The event has prob­a­bil­ity 0.

Answer: True.

Coins#

Ques­tion 5#

Two coins are tossed together. Find the prob­a­bil­ity of get­ting at most one tail.

The out­comes are HH, HT, TH, TT, so $n = 4$. "At most one tail" means no tail or one tail: HH, HT, TH, so $m = 3$.

$$\displaystyle P(\text{at most one tail}) = \frac{3}{4}.$$

Why this works: only TT has more than one tail, so the answer is also $\displaystyle 1 - \frac{1}{4}$.

Answer: $\displaystyle \frac{3}{4}$

Ques­tion 6#

Three coins are tossed. Find the prob­a­bil­ity of get­ting (a) at least one tail, (b) exactly one head, (c) all three coins show­ing the same face.

The 8 out­comes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

(a) Only HHH has no tail, so $\displaystyle P = 1 - \frac{1}{8} = \frac{7}{8}$.

(b) Exactly one head: HTT, THT, TTH, so $\displaystyle P = \frac{3}{8}$.

(c) Same face: HHH and TTT, so $\displaystyle P = \frac{2}{8} = \frac{1}{4}$.

Answer: (a) $\displaystyle \frac{7}{8}$ (b) $\displaystyle \frac{3}{8}$ (c) $\displaystyle \frac{1}{4}$

Dice#

Ques­tion 7#

A die is rolled once. Find the prob­a­bil­ity of get­ting (a) a fac­tor of 6, (b) a num­ber greater than 6.

(a) The fac­tors of 6 on a die are 1, 2, 3 and 6, so $\displaystyle P = \frac{4}{6} = \frac{2}{3}$.

(b) No face is greater than 6, so this is an impos­si­ble event: $P = 0$.

Answer: (a) $\displaystyle \frac{2}{3}$ (b) $0$

Ques­tion 8#

Two dice are rolled together. Find the prob­a­bil­ity of get­ting (a) a sum of 8, (b) a dou­blet of even num­bers.

There are 36 out­comes.

(a) Sum 8: $(2,6), (3,5), (4,4), (5,3), (6,2)$, so $\displaystyle P = \frac{5}{36}$.

(b) Even dou­blets: $(2,2), (4,4), (6,6)$, so $\displaystyle P = \frac{3}{36} = \frac{1}{12}$.

Answer: (a) $\displaystyle \frac{5}{36}$ (b) $\displaystyle \frac{1}{12}$

Ques­tion 9#

Two dice are rolled. Find the prob­a­bil­ity that the dif­fer­ence between the two num­bers is 2.

List the ordered pairs whose num­bers dif­fer by 2:

$$(1,3),\ (3,1),\ (2,4),\ (4,2),\ (3,5),\ (5,3),\ (4,6),\ (6,4).$$

That is 8 out­comes out of 36.

$$\displaystyle P(\text{difference } 2) = \frac{8}{36} = \frac{2}{9}.$$

Answer: $\displaystyle \frac{2}{9}$

Ques­tion 10#

In a game, Meena rolls two dice. She wins if the prod­uct of the two num­bers is a per­fect square. Find the prob­a­bil­ity that she (a) wins, (b) does not win.

The pos­si­ble per­fect-square prod­ucts are 1, 4, 9, 16, 25 and 36.

• Prod­uct 1: $(1,1)$.
• Prod­uct 4: $(1,4), (4,1), (2,2)$.
• Prod­uct 9: $(3,3)$.
• Prod­uct 16: $(4,4)$.
• Prod­uct 25: $(5,5)$.
• Prod­uct 36: $(6,6)$.

That is 8 favourable out­comes.

$$\displaystyle P(\text{win}) = \frac{8}{36} = \frac{2}{9}, \qquad P(\text{not win}) = 1 - \frac{2}{9} = \frac{7}{9}.$$

Why this works: every dou­blet gives a square prod­uct, and $(1,4)$, $(4,1)$ are the only non-dou­blets that do.

Answer: (a) $\displaystyle \frac{2}{9}$ (b) $\displaystyle \frac{7}{9}$

Ques­tion 11#

Two dice are rolled. Find the prob­a­bil­ity that (a) the sum is not 7, (b) the sum is greater than 9.

(a) Sum 7 occurs in 6 of the 36 out­comes, so $\displaystyle P(\text{sum } 7) = \frac{1}{6}$ and

$$\displaystyle P(\text{sum not } 7) = 1 - \frac{1}{6} = \frac{5}{6}.$$

(b) Sum 10: 3 ways; sum 11: 2 ways; sum 12: 1 way. Total 6, so $\displaystyle P = \frac{6}{36} = \frac{1}{6}$.

Answer: (a) $\displaystyle \frac{5}{6}$ (b) $\displaystyle \frac{1}{6}$

Cards#

Ques­tion 12#

A card is drawn at ran­dom from a well-shuf­fled pack of 52 cards. Find the prob­a­bil­ity that it is (a) the queen of dia­monds, (b) not a face card, (c) a black card that is not a king.

(a) There is exactly one queen of dia­monds: $\displaystyle P = \frac{1}{52}$.

(b) Cards that are not face cards: $52 - 12 = 40$, so $\displaystyle P = \frac{40}{52} = \frac{10}{13}$.

(c) There are 26 black cards, of which 2 are kings, leav­ing 24. So $\displaystyle P = \frac{24}{52} = \frac{6}{13}$.

Answer: (a) $\displaystyle \frac{1}{52}$ (b) $\displaystyle \frac{10}{13}$ (c) $\displaystyle \frac{6}{13}$

Ques­tion 13#

All the clubs are removed from a pack of 52 cards. The remain­ing cards are shuf­fled and one is drawn. Find the prob­a­bil­ity that it is (a) red, (b) a king, (c) a club, (d) a face card.

Cards left: $52 - 13 = 39$ (spades, hearts, dia­monds).

(a) All 26 red cards remain: $\displaystyle P = \frac{26}{39} = \frac{2}{3}$.

(b) Kings left: 3, so $\displaystyle P = \frac{3}{39} = \frac{1}{13}$.

(c) No clubs remain, so $P = 0$.

(d) Face cards left: $3 \times 3 = 9$, so $\displaystyle P = \frac{9}{39} = \frac{3}{13}$.

Answer: (a) $\displaystyle \frac{2}{3}$ (b) $\displaystyle \frac{1}{13}$ (c) $0$ (d) $\displaystyle \frac{3}{13}$

Balls, coins and num­bers#

Ques­tion 14#

A bag con­tains 4 red balls and some white balls. The prob­a­bil­ity of draw­ing a white ball is twice the prob­a­bil­ity of draw­ing a red ball. Find the num­ber of white balls.

Let there be $x$ white balls, so the total is $4 + x$.

$$\displaystyle \begin{aligned} P(\text{white}) &= 2 \times P(\text{red}) \\ \frac{x}{4 + x} &= 2 \times \frac{4}{4 + x} \\ x &= 8. \end{aligned}$$

Mul­ti­ply­ing both sides of the sec­ond line by $4 + x$ (which is pos­i­tive) gives $x = 8$.

Check: with 12 balls, $\displaystyle P(\text{white}) = \frac{8}{12} = \frac{2}{3}$ and $\displaystyle P(\text{red}) = \frac{4}{12} = \frac{1}{3}$; the first is twice the sec­ond and they add to 1.

Answer: 8 white balls

Ques­tion 15#

A money box con­tains sixty ₹1 coins, thirty ₹2 coins, twenty-five ₹5 coins and five ₹10 coins. One coin falls out at ran­dom when the box is turned upside down. Find the prob­a­bil­ity that the coin (a) is a ₹5 coin, (b) is not a ₹10 coin, (c) is worth less than ₹5.

Total coins: $60 + 30 + 25 + 5 = 120$.

(a) There are 25 ₹5 coins, so $\displaystyle P = \frac{25}{120} = \frac{5}{24}$.

(b) Coins that are not ₹10: $120 - 5 = 115$, so $\displaystyle P = \frac{115}{120} = \frac{23}{24}$.

(c) Coins worth less than ₹5 are the ₹1 and ₹2 coins: $60 + 30 = 90$, so $\displaystyle P = \frac{90}{120} = \frac{3}{4}$.

Answer: (a) $\displaystyle \frac{5}{24}$ (b) $\displaystyle \frac{23}{24}$ (c) $\displaystyle \frac{3}{4}$

Ques­tion 16#

An inte­ger $x$ is cho­sen at ran­dom from $-4, -3, -2, -1, 0, 1, 2, 3, 4$. Find the prob­a­bil­ity that (a) $x^2 \leq 4$, (b) $x^2 \gt 16$.

There are 9 equally likely val­ues.

(a) $x^2 \leq 4$ holds for $x = -2, -1, 0, 1, 2$, so $\displaystyle P = \frac{5}{9}$.

(b) The largest value of $x^2$ is 16, so $x^2 \gt 16$ never holds: $P = 0$.

Answer: (a) $\displaystyle \frac{5}{9}$ (b) $0$

Ques­tion 17#

A lot of 150 LED bulbs con­tains 120 good bulbs, 21 slightly dim bulbs and 9 faulty bulbs. Shop­keeper A buys a bulb only if it is good, while shop­keeper B buys a bulb unless it is faulty. One bulb is taken out at ran­dom. Find the prob­a­bil­ity that (a) A will buy it, (b) B will buy it, (c) it is faulty.

(a) A buys only the 120 good bulbs:

$$\displaystyle P(\text{A buys}) = \frac{120}{150} = \frac{4}{5}.$$

(b) B buys good or dim bulbs, $120 + 21 = 141$ of them:

$$\displaystyle P(\text{B buys}) = \frac{141}{150} = \frac{47}{50}.$$

(c) $\displaystyle P(\text{faulty}) = \frac{9}{150} = \frac{3}{50}$. As a check, "B buys" and "faulty" are com­ple­men­tary: $\displaystyle \frac{47}{50} + \frac{3}{50} = 1$.

Answer: (a) $\displaystyle \frac{4}{5}$ (b) $\displaystyle \frac{47}{50}$ (c) $\displaystyle \frac{3}{50}$

Ques­tion 18#

Cards num­bered 1 to 40 are placed in a box and one is drawn at ran­dom. Find the prob­a­bil­ity that the num­ber is (a) a prime num­ber, (b) not a mul­ti­ple of 3.

(a) Primes up to 40: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, which is 12 num­bers.

$$\displaystyle P(\text{prime}) = \frac{12}{40} = \frac{3}{10}.$$

(b) Mul­ti­ples of 3 up to 40: $3, 6, \ldots, 39$, which is 13 num­bers. So

$$\displaystyle P(\text{not a multiple of } 3) = 1 - \frac{13}{40} = \frac{27}{40}.$$

Answer: (a) $\displaystyle \frac{3}{10}$ (b) $\displaystyle \frac{27}{40}$

Key terms#

Equally likely

Out­comes that each have the same chance of occur­ring.

Favourable out­come

An out­come that makes the event in ques­tion hap­pen.

Impos­si­ble event

An event with prob­a­bil­ity 0, such as a sum of 13 on two dice.

Sure event

An event with prob­a­bil­ity 1.

Com­ple­men­tary events

An event and its "not" event, whose prob­a­bil­i­ties add to 1.

Dou­blet

An out­come of two dice with the same num­ber on both.

Face card

A jack, queen or king of any suit.

Com­mon ques­tions#

How much work­ing should I show for a one-mark ques­tion?#

Write the total num­ber of out­comes, the favourable out­comes, and the frac­tion. That short line is enough to earn the mark and to let you check your count.

Should I leave the answer as a frac­tion or a dec­i­mal?#

Give a frac­tion in low­est terms unless the ques­tion uses dec­i­mals. Both are cor­rect if they are equal.

How do I avoid miss­ing out­comes with two dice?#

Draw or imag­ine the 6 by 6 grid and go row by row. Remem­ber that $(a, b)$ and $(b, a)$ are sep­a­rate out­comes when $a \neq b$.

When is the com­ple­ment quicker?#

When­ever the ques­tion says "not", "at least one" or "does not win". Count the oppo­site event, which is usu­ally smaller, and sub­tract its prob­a­bil­ity from 1.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.