Tag: Exercise 14.1 Chapter 14 Probability Class 10th Maths NCERT Solutions

1. Complete the following statements:

(i) Probability of an event $E$ + Probability of the event ‘not $E$ = $\underset{‾}{1}$.

(ii) The probability of an event that cannot happen is $\underset{‾}{0}$. Such an event is called $\underset{‾}{\text{an impossible event}}$.

(iii) The probability of an event that is certain to happen is $\underset{‾}{1}$. Such an event is called $\underset{‾}{\text{a sure (certain) event}}$.

(iv) The sum of the probabilities of all the elementary events of an experiment is $\underset{‾}{1}$.

(v) The probability of an event is greater than or equal to $\underset{‾}{0}$ and less than or equal to $\underset{‾}{1}$.
(That is, $0\le P(E)\le 1$

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2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

Answer: Not necessarily equally likely. Starting depends on many factors (battery, fuel, etc.). We cannot assume equal chance.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer: Not necessarily equally likely. The probabilities depend on the player’s skill, distance, defence, etc.

(iii) A trial is made to answer a true–false question. The answer is right or wrong.

Answer: If the answer is chosen at random (a pure guess), the two outcomes (right/wrong) are equally likely (each probability $1\mathrm{/}2$). If the student knows the answer (or mostly knows), they are not equally likely. So: equally likely only when guessing at random.

(iv) A baby is born. It is a boy or a girl.

Answer: For practical school-problems we assume the two outcomes are approximately equally likely (each about $1\mathrm{/}2$). In reality there is a slight natural bias, but the experiment is normally treated as equally likely.

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3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution: A (fair) coin is symmetric and has two equally likely outcomes — Head or Tail. Thus each team has equal probability $1\mathrm{/}2$ of being chosen; this makes the choice unbiased and fair.

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4. Which of the following cannot be the probability of an event?
(A) $\frac{2}{3}$ (B) $-1.5$ (C) $15\mathrm{\%}$ (D) $0.7$

Solution: A $\frac{2}{3}$ is valid ($\approx 0.666$), C $15\mathrm{\%}=0.15$ valid, D $0.7$ valid. (B) $-1.5$ is impossible (probabilities cannot be negative).
Answer: (B) $-1.5$ cannot be a probability.

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5. If $P(E)=0.05$, what is the probability of ‘not $E$?

Solution: $P(\text{not }E)=1-P(E)=1-0.05=0.95$

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6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Solution: (i) Orange flavoured: impossible ⇒ probability $0$.
(ii) Lemon flavoured: certain ⇒ probability $1$.

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7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is $0.992$. What is the probability that the 2 students have the same birthday?

Solution: Probability same birthday $=1-0.992=\mathrm{0.008.}$

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8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Solution: Total $=3+5=8$
(i) $P(\text{red})={\displaystyle \frac{3}{8}}$
(ii) $P(\text{not red})={\displaystyle \frac{5}{8}}$

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9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

Solution: Total $=5+8+4=17$
(i) $P(\text{red})={\displaystyle \frac{5}{17}}$
(ii) $P(\text{white})={\displaystyle \frac{8}{17}}$
(iii) $P(\text{not green})=1-P(\text{green})=1-{\displaystyle \frac{4}{17}}={\displaystyle \frac{13}{17}}\mathrm{}$

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10. A piggy bank contains hundred 50p coins, fifty $\text{₹}1$ coins, twenty $\text{₹}2$ coins and ten $\text{₹}5$ coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a $\text{₹}5$ coin?

Solution: Total coins $=100+50+20+10=180$
(i) $P(\text{50p})={\displaystyle \frac{100}{180}}={\displaystyle \frac{10}{18}}={\displaystyle \frac{5}{9}}\mathrm{}$
(ii) $P(\text{not ₹5})=1-P(\text{₹5})=1-{\displaystyle \frac{10}{180}}=1-{\displaystyle \frac{1}{18}}={\displaystyle \frac{17}{18}}\mathrm{}$

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11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution: Total $=5+8=13$. So $P(\text{male})={\displaystyle \frac{5}{13}}\mathrm{}$

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12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1,2,\dots ,8$ (all equally likely). What is the probability that it will point at

(i) $8$?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution: Total outcomes $=8$
(i) $P(8)={\displaystyle \frac{1}{8}}\mathrm{.}$
(ii) Odd numbers $=1,3,5,7$ ⇒ $4$ outcomes ⇒ $P={\displaystyle \frac{4}{8}}={\displaystyle \frac{1}{2}}\mathrm{}$
(iii) Numbers $>2$ are $3,4,5,6,7,8$ ⇒ $6$ outcomes ⇒ $P={\displaystyle \frac{6}{8}}={\displaystyle \frac{3}{4}}\mathrm{}$
(iv) Numbers $<9$ are $1$ to $8$ ⇒ all outcomes ⇒ $P=1$

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13. A die is thrown once. Find the probability of getting

(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Solution: Faces $=1,2,3,4,5,6$ Total $=6$
(i) Primes: $2,3,5$ ⇒ $3$ outcomes ⇒ $P={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}\mathrm{.}$
(ii) “Lying between 2 and 6” (interpreted as strictly between): $3,4,5$ ⇒ $3$ outcomes ⇒ $P={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}\mathrm{}$

(iii) Odd numbers: $1,3,5$ ⇒ $3$ outcomes ⇒ $P={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}\mathrm{}$

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14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds

Solution: Deck has 52 cards; each equally likely. Face cards = J,Q,K (3 per suit), total 12.
(i) Red kings: two red suits (hearts, diamonds) ⇒ 2 kings ⇒ $P={\displaystyle \frac{2}{52}}={\displaystyle \frac{1}{26}}\mathrm{}$
(ii) Face card: $12$ cards ⇒ $P={\displaystyle \frac{12}{52}}={\displaystyle \frac{3}{13}}\mathrm{}$
(iii) Red face card: red suits have $3$ face cards each ⇒ $6$ cards ⇒ $P={\displaystyle \frac{6}{52}}={\displaystyle \frac{3}{26}}\mathrm{}$
(iv) Jack of hearts: $1$ card ⇒ $P={\displaystyle \frac{1}{52}}\mathrm{}$
(v) A spade: $13$ spades ⇒ $P={\displaystyle \frac{13}{52}}={\displaystyle \frac{1}{4}}\mathrm{}$
(vi) Queen of diamonds: $1$ card ⇒ $P={\displaystyle \frac{1}{52}}\mathrm{}$

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15. Five cards—the ten, jack, queen, king and ace of diamonds—are well-shuffled face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Solution: Initially 5 cards, all equally likely.
(i) $P(\text{queen})={\displaystyle \frac{1}{5}}\mathrm{}$
(ii) After queen removed, remaining cards $=4$ (ten, jack, king, ace).
(a) $P(\text{ace})={\displaystyle \frac{1}{4}}\mathrm{}$
(b) $P(\text{queen})=0$ (queen already removed).

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16. 12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random. Determine the probability that the pen taken out is a good one.

Solution: Total $=12+132=\mathrm{144<span\; class="katex"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord">.</span></span></span></span><span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">\; Good\; pens </span>}$$=132$
$P(\text{good})={\displaystyle \frac{132}{144}}={\displaystyle \frac{11}{12}}\approx \mathrm{0.9167.}$

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17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution: (i) $P(\text{defective})={\displaystyle \frac{4}{20}}={\displaystyle \frac{1}{5}}\mathrm{}$
(ii) Initially good bulbs $=20-4=16$. If the first drawn is not defective (i.e. a good one) and not replaced, remaining bulbs $=19$ and remaining good $=15$. So $P(\text{not defective second})={\displaystyle \frac{15}{19}}\mathrm{}$

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18. A box contains 90 discs numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5

Solution: Total outcomes $=90$

(i) Two-digit numbers from 10 to 90 inclusive ⇒ count = $90-10+1=81$
$P={\displaystyle \frac{81}{90}}={\displaystyle \frac{9}{10}}\mathrm{}$

(ii) Perfect squares $\le 90$: $1^2,2^2,\dots ,9^2=1,4,9,16,25,36,49,64,81$ ⇒ 9 numbers.
$P={\displaystyle \frac{9}{90}}={\displaystyle \frac{1}{10}}\mathrm{}$

(iii) Numbers divisible by 5 up to 90: $5,10,\dots ,90$. Count $=90\mathrm{/}5=18$
$P={\displaystyle \frac{18}{90}}={\displaystyle \frac{1}{5}}\mathrm{.}$

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19. A child has a die whose six faces show the letters as given below:

$ABCDEA$

The die is thrown once. What is the probability of getting (i) $A$? (ii) $D$?

Solution: There are 6 faces; $A$ appears twice.
(i) $P(A)={\displaystyle \frac{2}{6}}={\displaystyle \frac{1}{3}}\mathrm{}$
(ii) $D$ appears once ⇒ $P(D)={\displaystyle \frac{1}{6}}\mathrm{}$

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20. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter $1$m?* (Area method)

Solution: Rectangle area = $3\text{ m}\times 2\text{ m}=6{\text{ m}}^2$ (from figure labels). Circle radius $r=\frac{1}{2}$, area $=\pi r^2=\pi {\left(\frac{1}{2}\right)}^2={\displaystyle \frac{\pi }{4}}\mathrm{}$
Probability $={\displaystyle \frac{\text{area of circle}}{\text{area of rectangle}}}={\displaystyle \frac{\pi \mathrm{/}4}{6}}={\displaystyle \frac{\pi }{24}}\mathrm{}$(Approx.) ${\displaystyle \frac{\pi }{24}}\approx 0.1309$

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21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it ?
(ii) She will not buy it ?

Solution: Good $=144-20=124$ Total $=144.$
(i) $P(\text{buy})={\displaystyle \frac{124}{144}}={\displaystyle \frac{31}{36}}\mathrm{}$
(ii) $P(\text{not buy})={\displaystyle \frac{20}{144}}={\displaystyle \frac{5}{36}}\mathrm{}$

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22. Refer to Example 13 (two dice).

(i) Complete the following table:

Event ‘Sum on 2 dice’: $2,3,4,5,6,7,8,9,10,11,12$
Probability: ${\displaystyle \frac{1}{36}},{\displaystyle \frac{2}{36}},{\displaystyle \frac{3}{36}},{\displaystyle \frac{4}{36}},{\displaystyle \frac{5}{36}},{\displaystyle \frac{6}{36}},{\displaystyle \frac{5}{36}},{\displaystyle \frac{4}{36}},{\displaystyle \frac{3}{36}},{\displaystyle \frac{2}{36}},{\displaystyle \frac{1}{36}}$

(ii) A student argues that ‘there are 11 possible outcomes (2 through 12). Therefore, each has probability $1\mathrm{/}11$. Do you agree?

Solution: (i) Completed as above — probabilities come from counting ordered pairs $(i,j)$ with $i,j\in \{1,\dots ,6\}$ (ii) No — incorrect. The sums are not equally likely: e.g., sum 7 can occur in 6 ways $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ while sum 2 occurs only as $(1,1)$. There are 36 equally likely ordered outcomes, and each sum’s probability equals (number of ordered pairs giving that sum)/36, not $1\mathrm{/}11$.

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23. A game consists of tossing a one-rupee coin 3 times. Hanif wins if all the tosses give the same result (three heads or three tails), and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution: Total outcomes $=2^3=8$. Winning outcomes = {HHH, TTT} ⇒ 2 outcomes ⇒ $P(\text{win})={\displaystyle \frac{2}{8}}={\displaystyle \frac{1}{4}}$. Thus $P(\text{lose})=1-{\displaystyle \frac{1}{4}}={\displaystyle \frac{3}{4}}\mathrm{}$

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24. A die is thrown twice. What is the probability that

(i) $5$ will not come up either time?
(ii) $5$ will come up at least once?

Solution: Treat two throws as independent. Probability a single throw is not 5 is ${\displaystyle \frac{5}{6}}$.

(i) Neither time ⇒ ${\left({\displaystyle \frac{5}{6}}\right)}^2={\displaystyle \frac{25}{36}}\mathrm{}$

(ii) At least once $=1-$ probability of none $=1-{\displaystyle \frac{25}{36}}={\displaystyle \frac{11}{36}}\mathrm{}$

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25. Which of the following arguments are correct and which are not? Give reasons.

(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is $1\mathrm{/}3$.

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is $1\mathrm{/}2$.

Solution: (i) Incorrect. Two tossed coins produce 4 equally likely ordered outcomes: HH, HT, TH, TT. Grouping them gives three events {HH}, {HT or TH}, {TT}, but these grouped events are not equally likely: middle event (“one of each”) has probability $2\mathrm{/}4=1\mathrm{/}2$, while HH and TT each have probability $1\mathrm{/}4$. So the claim $1\mathrm{/}3$ is wrong.

(ii) Correct. For a fair die the three odd faces $\{1,3,5\}$ and three even faces $\{2,4,6\}$ are each 3 outcomes out of 6. So $P(\text{odd})={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$