Class 10 Maths Chapter 14 Probability Important Questions with Answers in English

Q. If $P(E)=0.72$, what is the probability of ‘not E’?

Answer: $P(\overline{E})=1-0.72=0.28$.

Q. What is the probability of an impossible event? Give one example of such an event.

Answer: The probability of an impossible event is $0$. Example: getting 7 on one throw of a die.

Q. A letter is chosen at random from the word MATHEMATICS. What is the probability that it is M?

Answer: MATHEMATICS has 11 letters, and M appears 2 times. $P(M)=\dfrac{2}{11}$.

Q. Can $\dfrac{7}{6}$ be the probability of an event? Give a reason.

Answer: No. A probability always lies between 0 and 1, and $\dfrac76$ is greater than 1.

Q. A bag contains only white balls. One ball is drawn at random. What is the probability that it is white?

Answer: Every ball is white, so the event is sure: $P=1$.

Q. Two coins are tossed together. What is the probability of getting two tails?

Answer: Outcomes HH, HT, TH, TT; only TT is favourable. $P=\dfrac14$.

Q. One card is drawn from a well-shuffled deck of 52 cards. What is the probability that it is the queen of hearts?

Answer: There is exactly one queen of hearts. $P=\dfrac{1}{52}$.

Q. According to the chapter’s Note to the Reader, what may we expect about the experimental and theoretical probabilities of an event as the number of trials goes on increasing?

Answer: As the number of trials in an experiment goes on increasing, we may expect the experimental and theoretical probabilities to be nearly the same.

Q. What is an elementary event? Give an example.

Answer: An event having only one outcome of the experiment is called an elementary event. Example: getting a head when a coin is tossed once.

Q. A die is thrown once. Find the probability of getting (i) an even number (ii) a number at most 2.

Answer: Possible outcomes $=6$.
(i) Even numbers: 2, 4, 6. $P=\dfrac36=\dfrac12$.
(ii) At most 2 means 1 or 2. $P=\dfrac26=\dfrac13$.

Q. A bag contains 6 red, 4 green and 5 yellow balls. One ball is drawn at random. Find the probability that it is not green.

Answer: Total balls $=6+4+5=15$.
$P(\text{green})=\dfrac{4}{15}$, so $P(\text{not green})=1-\dfrac{4}{15}=\dfrac{11}{15}$.

Q. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a black face card (ii) a red ace.

Answer: (i) Black face cards: K, Q, J of spades and of clubs, so 6. $P=\dfrac{6}{52}=\dfrac{3}{26}$.
(ii) Red aces: ace of hearts and ace of diamonds, so 2. $P=\dfrac{2}{52}=\dfrac{1}{26}$.

Q. A number is chosen at random from the numbers 1 to 30. Find the probability that it is a prime number.

Answer: Primes from 1 to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, so 10.
$P=\dfrac{10}{30}=\dfrac13$.

Q. The probability of getting a bad egg from a crate of 400 eggs is 0.035. How many bad eggs are there in the crate?

Answer: $P(\text{bad egg})=\dfrac{\text{number of bad eggs}}{400}=0.035$.
Number of bad eggs $=0.035\times400=14$.

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

Answer: Outcomes: HH, HT, TH, TT (4, equally likely).
At most one head means 0 or 1 head: HT, TH, TT, so 3.
$P=\dfrac34$.

Q. When two dice are thrown, the sum can be any of the 11 numbers 2, 3, …, 12. Explain why the probability of each sum is not $\dfrac{1}{11}$.

Answer: The formula needs equally likely outcomes, and the 11 sums are not equally likely. The equally likely outcomes are the 36 ordered pairs. Sum 2 comes only from (1, 1), so $P(2)=\dfrac{1}{36}$, while sum 7 comes from 6 pairs, so $P(7)=\dfrac{6}{36}=\dfrac16$.

Q. A box has 25 cards numbered 1 to 25. One card is drawn at random. Find the probability that the number is divisible by both 2 and 3.

Answer: Divisible by both 2 and 3 means divisible by 6: 6, 12, 18, 24, so 4.
$P=\dfrac{4}{25}$.

Q. Asha and Bina are friends. Take a year of 365 days (not a leap year), with every day equally likely to be a birthday. Find the probability that (i) their birthdays fall on different days (ii) they share the same birthday.

Answer: Asha’s birthday can be any day of the year. Bina’s birthday can be any of the 365 days, and we assume these 365 outcomes are equally likely.
(i) Bina’s birthday is different from Asha’s on $365-1=364$ days. $P(\text{different})=\dfrac{364}{365}$.
(ii) ‘Same birthday’ is the complement of ‘different birthdays’. $P(\text{same})=1-\dfrac{364}{365}=\dfrac{1}{365}$.

Q. Two dice are thrown together. Find the probability that (i) both dice show the same number (ii) the sum is 9 (iii) the sum is 10 or more (iv) the product of the numbers is less than 5.

Answer: Total outcomes $=6\times6=36$.
(i) (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6): 6 outcomes. $P=\dfrac{6}{36}=\dfrac16$.
(ii) Sum 9: (3, 6), (4, 5), (5, 4), (6, 3): 4 outcomes. $P=\dfrac{4}{36}=\dfrac19$.
(iii) Sum 10: 3 ways, sum 11: 2 ways, sum 12: 1 way, total 6. $P=\dfrac{6}{36}=\dfrac16$.
(iv) Product 1, 2, 3 or 4: (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 2): 8 outcomes. $P=\dfrac{8}{36}=\dfrac29$.

Q. Three coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at most two heads (iv) no head.

Answer: The 8 equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
(i) Exactly two heads: HHT, HTH, THH. $P=\dfrac38$.
(ii) At least two heads: the 3 above and HHH, so 4. $P=\dfrac48=\dfrac12$.
(iii) At most two heads: everything except HHH, so 7. $P=\dfrac78$.
(iv) No head: TTT only. $P=\dfrac18$.

Q. One card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is (i) neither a king nor a queen (ii) a red card that is not a face card (iii) either a club or a face card (iv) a number card (2 to 10) of spades.

Answer: Total outcomes $=52$.
(i) Kings and queens: $4+4=8$. Favourable $=52-8=44$. $P=\dfrac{44}{52}=\dfrac{11}{13}$.
(ii) Red cards 26, red face cards 6. Favourable $=26-6=20$. $P=\dfrac{20}{52}=\dfrac{5}{13}$.
(iii) Clubs 13, face cards 12, but the 3 club face cards are counted twice. Favourable $=13+12-3=22$. $P=\dfrac{22}{52}=\dfrac{11}{26}$.
(iv) Spades 2, 3, …, 10: 9 cards. $P=\dfrac{9}{52}$.

Q. A box contains 100 discs numbered 1 to 100. One disc is drawn at random. Find the probability that it bears (i) a two-digit number (ii) a perfect square (iii) a multiple of 7 (iv) a number divisible by both 4 and 6.

Answer: Total outcomes $=100$.
(i) Two-digit numbers 10 to 99: 90. $P=\dfrac{90}{100}=\dfrac{9}{10}$.
(ii) Perfect squares $1^2, 2^2, \ldots, 10^2$: 10. $P=\dfrac{10}{100}=\dfrac{1}{10}$.
(iii) Multiples of 7: 7, 14, …, 98, that is $7\times1$ to $7\times14$: 14. $P=\dfrac{14}{100}=\dfrac{7}{50}$.
(iv) Divisible by 4 and 6 means divisible by 12 (their LCM): 12, 24, …, 96: 8. $P=\dfrac{8}{100}=\dfrac{2}{25}$.

Q. A die is thrown twice. Find the probability that (i) 6 does not come up either time (ii) 6 comes up at least once (iii) the first throw gives an even number and the second an odd number (iv) the sum of the two numbers is odd.

Answer: Throwing a die twice is the same as throwing two dice: 36 outcomes.
(i) Each throw has 5 choices other than 6: $5\times5=25$. $P=\dfrac{25}{36}$.
(ii) Complement of (i): $P=1-\dfrac{25}{36}=\dfrac{11}{36}$.
(iii) First even (3 choices) and second odd (3 choices): $3\times3=9$. $P=\dfrac{9}{36}=\dfrac14$.
(iv) An odd sum needs one even and one odd number: $9$ (even, odd) $+\,9$ (odd, even) $=18$. $P=\dfrac{18}{36}=\dfrac12$.

Q. A bag contains 6 red balls and some blue balls. The probability of drawing a blue ball is twice the probability of drawing a red ball. (i) Find the number of blue balls. (ii) Find the probability of drawing a red ball.

Answer: (i) Let the number of blue balls be $x$. Total $=6+x$.
$P(\text{blue})=\dfrac{x}{6+x}$ and $P(\text{red})=\dfrac{6}{6+x}$.
Given $\dfrac{x}{6+x}=2\times\dfrac{6}{6+x}$, so $x=12$. There are 12 blue balls.
(ii) Total $=18$, so $P(\text{red})=\dfrac{6}{18}=\dfrac13$. (Check: $P(\text{blue})=\dfrac{12}{18}=\dfrac23$, which is twice $\dfrac13$.)

Q. Cards numbered 2 to 101 are placed in a box and mixed well. One card is drawn at random. Find the probability that the number on it is (i) an even number (ii) a number less than 14 (iii) a perfect square (iv) a prime number less than 20.

Answer: Numbers 2 to 101: $101-2+1=100$ cards.
(i) Even numbers 2, 4, …, 100: 50. $P=\dfrac{50}{100}=\dfrac12$.
(ii) Numbers 2 to 13: 12. $P=\dfrac{12}{100}=\dfrac{3}{25}$.
(iii) Perfect squares $2^2, 3^2, \ldots, 10^2$ (1 is not in the box): 9. $P=\dfrac{9}{100}$.
(iv) Primes below 20: 2, 3, 5, 7, 11, 13, 17, 19: 8. $P=\dfrac{8}{100}=\dfrac{2}{25}$.

Q. A packet of 24 bulbs contains 6 defective bulbs. (i) One bulb is drawn at random. Find the probability that it is defective. (ii) The bulb drawn in (i) is defective and is not replaced. Find the probability that the next bulb drawn is not defective. (iii) If instead the first bulb drawn was good and not replaced, find the probability that the next bulb drawn is defective.

Answer: (i) $P(\text{defective})=\dfrac{6}{24}=\dfrac14$.
(ii) After removing one defective bulb: 23 bulbs left, of which $24-6=18$ are good. $P=\dfrac{18}{23}$.
(iii) After removing one good bulb: 23 bulbs left, still 6 defective. $P=\dfrac{6}{23}$.

Q. Two dice, one red and one blue, are thrown together. Write down the number of possible outcomes and find the probability that (i) both dice show the same even number (ii) the sum is a perfect square (iii) the sum is a prime number (iv) the number on the red die is greater than the number on the blue die (v) the two numbers differ by 2.

Answer: Each outcome is an ordered pair (red, blue): $6\times6=36$ equally likely outcomes.
(i) (2, 2), (4, 4), (6, 6): 3. $P=\dfrac{3}{36}=\dfrac{1}{12}$.
(ii) Possible perfect-square sums are 4 and 9. Sum 4: (1, 3), (2, 2), (3, 1); sum 9: (3, 6), (4, 5), (5, 4), (6, 3). Total 7. $P=\dfrac{7}{36}$.
(iii) Prime sums 2, 3, 5, 7, 11 occur in $1+2+4+6+2=15$ ways. $P=\dfrac{15}{36}=\dfrac{5}{12}$.
(iv) Of the 36 pairs, 6 are doublets; the remaining 30 split equally into red greater and blue greater. So 15. $P=\dfrac{15}{36}=\dfrac{5}{12}$.
(v) (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4): 8. $P=\dfrac{8}{36}=\dfrac29$.

Q. A money box contains 120 coins: fifty ₹1 coins, forty ₹2 coins, twenty ₹5 coins and ten ₹10 coins. It is equally likely that any one coin falls out when the box is turned upside down. Find the probability that the coin that falls out (i) is a ₹5 coin (ii) is not a ₹1 coin (iii) has a value of at least ₹5 (iv) has a value less than ₹10. Also verify that the probabilities of the four kinds of coin add up to 1.

Answer: Total coins $=50+40+20+10=120$.
(i) P(₹5 coin) $=\dfrac{20}{120}=\dfrac16$.
(ii) P(₹1 coin) $=\dfrac{50}{120}=\dfrac{5}{12}$, so P(not a ₹1 coin) $=1-\dfrac{5}{12}=\dfrac{7}{12}$.
(iii) Value at least ₹5 means ₹5 or ₹10: $20+10=30$. $P=\dfrac{30}{120}=\dfrac14$.
(iv) Value less than ₹10 means every coin except ₹10: $120-10=110$. $P=\dfrac{110}{120}=\dfrac{11}{12}$.
Check: $\dfrac{50}{120}+\dfrac{40}{120}+\dfrac{20}{120}+\dfrac{10}{120}=\dfrac{120}{120}=1$.

Q. Define each of the following with one example: (i) equally likely outcomes (ii) theoretical probability (iii) elementary event (iv) impossible event (v) sure event (vi) complementary events. Also state the range of the probability of an event.

Answer: (i) Equally likely outcomes: outcomes that each have the same chance of happening. Example: head and tail on a fair coin.
(ii) Theoretical probability: $P(E)=\dfrac{\text{number of outcomes favourable to } E}{\text{number of all possible outcomes}}$, assuming equally likely outcomes. Example: $P(\text{head})=\dfrac12$.
(iii) Elementary event: an event having only one outcome of the experiment. Example: getting 5 on one throw of a die.
(iv) Impossible event: an event that cannot happen; its probability is 0. Example: getting 8 on one throw of a die.
(v) Sure (certain) event: an event that is certain to happen; its probability is 1. Example: getting a number less than 7 on one throw of a die.
(vi) Complementary events: E and ‘not E’ ($\overline{E}$); $P(E)+P(\overline{E})=1$. Example: getting a head and getting a tail on one toss.
Range: $0\le P(E)\le1$ for every event E.

Q. A game of chance consists of spinning an arrow which comes to rest at one of the numbers 1 to 12, all equally likely. Find the probability that it points at (i) 7 (ii) an odd number (iii) a number greater than 9 (iv) a prime number (v) a multiple of 3 (vi) a number less than 13.

Answer: Total outcomes $=12$.
(i) $P(7)=\dfrac{1}{12}$.
(ii) Odd: 1, 3, 5, 7, 9, 11, so 6. $P=\dfrac{6}{12}=\dfrac12$.
(iii) Greater than 9: 10, 11, 12. $P=\dfrac{3}{12}=\dfrac14$.
(iv) Primes: 2, 3, 5, 7, 11, so 5. $P=\dfrac{5}{12}$.
(v) Multiples of 3: 3, 6, 9, 12. $P=\dfrac{4}{12}=\dfrac13$.
(vi) Every number is less than 13: sure event, $P=\dfrac{12}{12}=1$.