Class 10 Probability Problems: Coins, Dice and Playing Cards

List­ing with a tree dia­gram#

A tree dia­gram makes sure that noth­ing is missed. Each level of branches stands for one coin, and each path from the start to the end is one out­come.

Words to read care­fully#

• Exactly two heads: two heads and no more.
• At least two heads: two or more heads.
• At most one head: one head or none.
• No tail: every coin shows a head.

Dice#

One die#

One die has 6 out­comes: 1, 2, 3, 4, 5, 6. Among them, 2, 3 and 5 are prime; 4 is the only com­pos­ite num­ber; 1 is nei­ther prime nor com­pos­ite.

Two dice: the 36-out­come grid#

When two dice are rolled (or one die is rolled twice), each of the 6 results of the first die can pair with each of the 6 results of the sec­ond, so there are $6 \times 6 = 36$ out­comes. Write each as an ordered pair $(a, b)$, where $a$ is on the first die and $b$ on the sec­ond. $(2, 5)$ and $(5, 2)$ are dif­fer­ent out­comes.

Use­ful counts from the grid:

The counts add up to 36. A dou­blet is an out­come with the same num­ber on both dice; there are 6 of them.

A pack of cards#

Struc­ture of the pack#

A stan­dard pack has 52 cards in 4 suits of 13 cards each. Spades and clubs are black; hearts and dia­monds are red. Each suit has an ace, the num­ber cards 2 to 10, and three face cards: the jack, the queen and the king.

• Red cards: 26; black cards: 26.
• Aces: 4; kings: 4; queens: 4; jacks: 4.
• Face cards: $3 \times 4 = 12$.
• Cards of any one suit: 13.

Method for any count­ing prob­lem#

1. Iden­tify the exper­i­ment and write down $n$: 4 or 8 for coins, 6 or 36 for dice, 52 (or fewer, if cards are removed) for a pack.
2. Trans­late the event into a pre­cise list of out­comes; draw a tree or grid if needed.
3. Count the favourable out­comes, avoid­ing dou­ble count­ing (for exam­ple, a card that is both a heart and a jack).
4. Divide, sim­plify, and check that the answer lies between 0 and 1.

Worked exam­ples#

Exam­ple 1: two coins#

Two coins are tossed together. Find the prob­a­bil­ity of get­ting (a) at least one head, (b) exactly one head, (c) no head.

The out­comes are HH, HT, TH, TT, so $n = 4$.

(a) At least one head: HH, HT, TH, so $\displaystyle P = \frac{3}{4}$.

(b) Exactly one head: HT, TH, so $\displaystyle P = \frac{2}{4} = \frac{1}{2}$.

(c) No head: TT, so $\displaystyle P = \frac{1}{4}$. Notice that (a) and (c) are com­ple­men­tary: $\displaystyle \frac{3}{4} + \frac{1}{4} = 1$.

Answer: (a) $\displaystyle \frac{3}{4}$ (b) $\displaystyle \frac{1}{2}$ (c) $\displaystyle \frac{1}{4}$

Exam­ple 2: three coins#

Three coins are tossed. Find the prob­a­bil­ity of get­ting (a) exactly two heads, (b) at least two heads, (c) no tail, (d) at least one tail.

Here $n = 8$.

(a) HHT, HTH, THH, so $\displaystyle P = \frac{3}{8}$.

(b) The three out­comes in (a) together with HHH, so $\displaystyle P = \frac{4}{8} = \frac{1}{2}$.

(c) Only HHH, so $\displaystyle P = \frac{1}{8}$.

(d) "At least one tail" is the com­ple­ment of "no tail", so $\displaystyle P = 1 - \frac{1}{8} = \frac{7}{8}$.

Answer: (a) $\displaystyle \frac{3}{8}$ (b) $\displaystyle \frac{1}{2}$ (c) $\displaystyle \frac{1}{8}$ (d) $\displaystyle \frac{7}{8}$

Exam­ple 3: sums and prod­ucts on two dice#

Two dice are rolled. Find the prob­a­bil­ity that (a) the sum is 7, (b) a dou­blet appears, (c) the sum is at most 4, (d) the prod­uct is 12, (e) the sum is a prime num­ber.

Here $n = 36$.

(a) $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$: $\displaystyle P = \frac{6}{36} = \frac{1}{6}$.

(b) $(1,1)$ to $(6,6)$: $\displaystyle P = \frac{6}{36} = \frac{1}{6}$.

(c) Sum 2, 3 or 4: $1 + 2 + 3 = 6$ out­comes, so $\displaystyle P = \frac{6}{36} = \frac{1}{6}$.

(d) $(2,6), (6,2), (3,4), (4,3)$: $\displaystyle P = \frac{4}{36} = \frac{1}{9}$.

(e) The pos­si­ble prime sums are 2, 3, 5, 7 and 11. From the table they occur in $1 + 2 + 4 + 6 + 2 = 15$ ways, so

$$\displaystyle P(\text{prime sum}) = \frac{15}{36} = \frac{5}{12}.$$

Answer: (a) $\displaystyle \frac{1}{6}$ (b) $\displaystyle \frac{1}{6}$ (c) $\displaystyle \frac{1}{6}$ (d) $\displaystyle \frac{1}{9}$ (e) $\displaystyle \frac{5}{12}$

Exam­ple 4: draw­ing one card#

One card is drawn from a well-shuf­fled pack of 52. Find the prob­a­bil­ity that it is (a) a red king, (b) a face card, (c) nei­ther a heart nor a jack, (d) a black ace.

(a) King of hearts and king of dia­monds: $\displaystyle P = \frac{2}{52} = \frac{1}{26}$.

(b) 12 face cards: $\displaystyle P = \frac{12}{52} = \frac{3}{13}$.

(c) First count "a heart or a jack": 13 hearts, plus the 3 jacks that are not hearts, giv­ing 16 (the jack of hearts is counted once only). So

$$\displaystyle P(\text{neither}) = \frac{52 - 16}{52} = \frac{36}{52} = \frac{9}{13}.$$

(d) Ace of spades and ace of clubs: $\displaystyle P = \frac{2}{52} = \frac{1}{26}$.

Answer: (a) $\displaystyle \frac{1}{26}$ (b) $\displaystyle \frac{3}{13}$ (c) $\displaystyle \frac{9}{13}$ (d) $\displaystyle \frac{1}{26}$

Exam­ple 5: after remov­ing cards#

The king, queen and jack of spades are removed from a pack of 52 cards. The rest are shuf­fled and one card is drawn. Find the prob­a­bil­ity that it is (a) a spade, (b) a face card, (c) an ace, (d) a red card.

Cards left: $n = 52 - 3 = 49$.

(a) Spades left: $13 - 3 = 10$, so $\displaystyle P = \frac{10}{49}$.

(b) Face cards left: $12 - 3 = 9$, so $\displaystyle P = \frac{9}{49}$.

(c) No ace was removed, so $\displaystyle P = \frac{4}{49}$.

(d) All 26 red cards remain, so $\displaystyle P = \frac{26}{49}$.

Answer: (a) $\displaystyle \frac{10}{49}$ (b) $\displaystyle \frac{9}{49}$ (c) $\displaystyle \frac{4}{49}$ (d) $\displaystyle \frac{26}{49}$

Exam­ple 6: find­ing an unknown num­ber of balls#

A bag con­tains 8 red balls and some blue balls. The prob­a­bil­ity of draw­ing a blue ball at ran­dom is $\displaystyle \frac{3}{5}$. Find the num­ber of blue balls.

Let there be $x$ blue balls. The total is $8 + x$, so

$$\displaystyle \begin{aligned} \frac{x}{8 + x} &= \frac{3}{5} \\ 5x &= 24 + 3x \\ 2x &= 24 \\ x &= 12. \end{aligned}$$

Check: $\displaystyle \frac{12}{20} = \frac{3}{5}$, and $\displaystyle P(\text{red}) = \frac{8}{20} = \frac{2}{5}$, which adds with $\displaystyle \frac{3}{5}$ to 1.

Answer: 12 blue balls

Exam­ple 7: num­bered discs#

A box has 90 discs num­bered 1 to 90. One disc is drawn at ran­dom. Find the prob­a­bil­ity that it bears (a) a two-digit num­ber, (b) a per­fect square, (c) a mul­ti­ple of 5.

(a) Two-digit num­bers: 10 to 90, which is $90 - 9 = 81$ num­bers. $\displaystyle P = \frac{81}{90} = \frac{9}{10}$.

(b) Per­fect squares up to 90: $1, 4, 9, 16, 25, 36, 49, 64, 81$, which is 9 num­bers. $\displaystyle P = \frac{9}{90} = \frac{1}{10}$.

(c) Mul­ti­ples of 5: $5, 10, \ldots, 90$, which is $90 \div 5 = 18$ num­bers. $\displaystyle P = \frac{18}{90} = \frac{1}{5}$.

Answer: (a) $\displaystyle \frac{9}{10}$ (b) $\displaystyle \frac{1}{10}$ (c) $\displaystyle \frac{1}{5}$

Exam­ple 8: 53 Sun­days in a leap year#

Find the prob­a­bil­ity that a leap year cho­sen at ran­dom has 53 Sun­days.

A leap year has 366 days, and $366 = 52 \times 7 + 2$. So it has 52 com­plete weeks, which give 52 Sun­days, plus 2 extra con­sec­u­tive days. The two extra days can be any of these 7 equally likely pairs:

(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun).

A 53rd Sun­day occurs when a Sun­day is among the two extra days, which hap­pens for 2 of the 7 pairs.

$$\displaystyle P(53 \text{ Sundays}) = \frac{2}{7}.$$

For an ordi­nary year, $365 = 52 \times 7 + 1$ leaves one extra day, so the prob­a­bil­ity of 53 Sun­days is $\displaystyle \frac{1}{7}$.

Answer: $\displaystyle \frac{2}{7}$

Com­mon mis­takes#

• Treat­ing $(2, 5)$ and $(5, 2)$ as the same out­come. With two dice there are 36 ordered out­comes, not 21.
• Count­ing a card twice, such as the jack of hearts in "a heart or a jack".
• For­get­ting to reduce the total after cards are removed from the pack.
• Call­ing the ace a face card. Only the jack, queen and king are face cards.
• Read­ing "at least one" as "exactly one". It is usu­ally eas­i­est as $1 - P(\text{none})$.
• In leap-year prob­lems, for­get­ting that the two extra days are con­sec­u­tive.

Try these#

1. Three coins are tossed. Find the prob­a­bil­ity of get­ting at most one head. Answer: $\displaystyle \frac{1}{2}$
2. Two dice are rolled. Find the prob­a­bil­ity that the sum is 10. Answer: $\displaystyle \frac{1}{12}$
3. Two dice are rolled. Find the prob­a­bil­ity that the sum is at least 10. Answer: $\displaystyle \frac{1}{6}$
4. A card is drawn from a pack of 52. Find the prob­a­bil­ity that it is a red face card. Answer: $\displaystyle \frac{3}{26}$
5. Find the prob­a­bil­ity that an ordi­nary (non-leap) year has 53 Mon­days. Answer: $\displaystyle \frac{1}{7}$
6. Two dice are rolled. Find the prob­a­bil­ity that the prod­uct of the num­bers is odd. Answer: $\displaystyle \frac{1}{4}$

Key terms#

Tree dia­gram

A branch­ing dia­gram in which every path from start to fin­ish is one out­come.

Ordered pair

An out­come such as $(a, b)$ in which the order mat­ters; $a$ belongs to the first die.

Dou­blet

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

Suit

One of the four groups of 13 cards: spades, clubs, hearts, dia­monds.

Face card

A jack, queen or king; a pack has 12 of them.

At least

That many or more.

At most

That many or fewer.

Com­mon ques­tions#

Is toss­ing one coin twice the same as toss­ing two coins once?#

Yes. Both have the same 4 equally likely out­comes, HH, HT, TH and TT.

Why are there 36 out­comes with two dice and not 21?#

The dice are dif­fer­ent objects, so a 2 on the first die with a 5 on the sec­ond is a dif­fer­ent out­come from a 5 on the first with a 2 on the sec­ond. Only the 36 ordered out­comes are equally likely.

Is the ace a face card?#

No. The face cards are the jack, queen and king of each suit, 12 cards in all.

How do I han­dle "at least one" ques­tions quickly?#

Find the prob­a­bil­ity of "none" and sub­tract it from 1. For three coins, $\displaystyle P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}$.

What changes when some cards are removed?#

Both the total and the favourable count may change. Recount each from the cards that are actu­ally left.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.