Ch 14: Probability (Dice & Cards)
Tackle the trickiest Probability questions involving Playing Cards, Two Dice, and Defective items (Ex 14.1 Q13-25 & Ex 14.2).
Exercise 14.1 (Questions 13 to 19)
💡 Trick for Cards: Total cards = 52. Suits = 4 (Spades, Hearts, Diamonds, Clubs). Face Cards = 12 (J, Q, K of all suits).
Q13. 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.
Total outcomes on a die = {1, 2, 3, 4, 5, 6} = 6.
(i) Prime numbers = {2, 3, 5} = 3 outcomes. P(prime) = 3/6 = 1/2.
(ii) Numbers between 2 and 6 = {3, 4, 5} = 3 outcomes. P(between 2 & 6) = 3/6 = 1/2.
(iii) Odd numbers = {1, 3, 5} = 3 outcomes. P(odd) = 3/6 = 1/2.
Q14. 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.
Total possible outcomes = 52.
(i) Kings of red colour (Hearts & Diamonds) = 2. P = 2/52 = 1/26.
(ii) Face cards (J, Q, K of 4 suits) = 12. P = 12/52 = 3/13.
(iii) Red face cards (Hearts & Diamonds) = 6. P = 6/52 = 3/26.
(iv) Jack of hearts = 1. P = 1/52.
(v) Spades = 13. P = 13/52 = 1/4.
(vi) Queen of diamonds = 1. P = 1/52.
Q15. Five cards (ten, jack, queen, king, ace of diamonds) are well-shuffled. One is picked. (i) Probability it is the queen? (ii) If the queen is drawn and put aside, what is the probability the second card is an ace? a queen?
(i) Total cards = 5. P(Queen) = 1/5.
(ii) If the queen is put aside, total remaining cards = 4.
(a) P(Ace) = 1/4.
(b) P(Queen) = 0 (since it is already removed).
Q16. 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.
Total pens = 12 + 132 = 144. Favorable outcomes (good pens) = 132.
P(good pen) = 132 / 144 = 11/12.
Q17. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn. Probability it is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn. Probability it is not defective?
(i) Total bulbs = 20. Defective = 4. P(defective) = 4 / 20 = 1/5.
(ii) Since a non-defective bulb is removed, total bulbs left = 19. Remaining good bulbs = 16 - 1 = 15.
P(not defective) = 15 / 19.
Q18. A box contains 90 discs numbered 1 to 90. Find the probability that a drawn disc bears: (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Total discs = 90.
(i) Two-digit numbers (10 to 90) = 81. P = 81/90 = 9/10.
(ii) Perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81) = 9. P = 9/90 = 1/10.
(iii) Divisible by 5 (5, 10, ... 90) = 18. P = 18/90 = 1/5.
Q19. A child has a die with letters A, B, C, D, E, A. Probability of getting: (i) A? (ii) D?
Total faces = 6. (i) Face A appears 2 times. P(A) = 2/6 = 1/3.
(ii) Face D appears 1 time. P(D) = 1/6.
Exercise 14.1 (Questions 20 to 25)
Q20. Suppose you drop a die at random on a rectangular region 3m by 2m. What is the probability that it will land inside the circle of diameter 1m?
Solution:
Total Area (Rectangle) = L $\times$ B = 3 $\times$ 2 = 6 m².
Favorable Area (Circle) = $\pi r^2 = \pi \times (1/2)^2 = \pi / 4$ m².
Probability = Favorable Area / Total Area = $(\pi / 4) / 6 =$ $\pi / 24$.
Q21. A lot consists of 144 ball pens of which 20 are defective. Nuri will buy a pen if it is good. Shopkeeper draws one. Probability that: (i) She will buy it? (ii) She will not buy it?
Total pens = 144. Defective = 20. Good pens = 144 - 20 = 124.
(i) P(buy it) = P(good pen) = 124 / 144 = 31/36.
(ii) P(not buy it) = P(defective pen) = 20 / 144 = 5/36.
Q22. Two dice are thrown. Complete the table for the sum on two dice (2 to 12).
Solution:
Total possible outcomes for 2 dice = 6 $\times$ 6 = 36.
- Sum 2: (1,1) $\rightarrow$ P = 1/36
- Sum 3: (1,2), (2,1) $\rightarrow$ P = 2/36
- Sum 4: (1,3), (2,2), (3,1) $\rightarrow$ P = 3/36
- Sum 5: (1,4), (2,3), (3,2), (4,1) $\rightarrow$ P = 4/36
- Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) $\rightarrow$ P = 5/36
- Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $\rightarrow$ P = 6/36
- Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) $\rightarrow$ P = 5/36
- Sum 9: (3,6), (4,5), (5,4), (6,3) $\rightarrow$ P = 4/36
- Sum 10: (4,6), (5,5), (6,4) $\rightarrow$ P = 3/36
- Sum 11: (5,6), (6,5) $\rightarrow$ P = 2/36
- Sum 12: (6,6) $\rightarrow$ P = 1/36
Q23. A game consists of tossing a one rupee coin 3 times. Hanif wins if all tosses give the same result (three heads or three tails). Calculate the probability that Hanif will lose.
Total outcomes for 3 coins = 2 $\times$ 2 $\times$ 2 = 8. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Winning outcomes (same result) = {HHH, TTT} = 2.
Losing outcomes = 8 - 2 = 6.
P(lose) = 6/8 = 3/4.
Q24. 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?
Total outcomes = 36.
Outcomes where 5 comes up at least once = (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6) $\rightarrow$ Total 11 outcomes.
(i) P(5 will not come up) = (36 - 11) / 36 = 25/36.
(ii) P(5 comes up at least once) = 11/36.
Q25. Which arguments are correct? (i) If two coins are tossed, there are three possible outcomes (two heads, two tails, one of each). So prob is 1/3. (ii) If a die is thrown, there are two outcomes: odd or even. So P(odd) = 1/2.
(i) Incorrect. The outcomes are {HH, HT, TH, TT}. Getting one of each has 2 outcomes (HT, TH). The events are not equally likely.
(ii) Correct. Odd numbers {1,3,5} and Even numbers {2,4,6}. Both have 3 outcomes. P(odd) = 3/6 = 1/2.
Exercise 14.2 (Optional)
Q1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tue to Sat). Each is equally likely to visit the shop on any day. What is the probability that they will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Solution:
Total days = 5 (Tue, Wed, Thu, Fri, Sat). Total outcomes = 5 $\times$ 5 = 25.
(i) Same day: (Tue,Tue), (Wed,Wed)... = 5 outcomes. P = 5/25 = 1/5.
(ii) Consecutive days: (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat) and reverse = 8 outcomes. P = 8/25.
(iii) Different days: 1 - P(same day) = 1 - 1/5 = 4/5.
Q2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times. Find the probability that the total sum is (i) even (ii) 6 (iii) at least 6.
Solution:
Total outcomes = 6 $\times$ 6 = 36. After drawing a table for the sums:
(i) Number of even sums (2,4,4,6,6,8,12) = 18. P = 18/36 = 1/2.
(ii) Number of sums equal to 6 = 4. P = 4/36 = 1/9.
(iii) Number of sums at least 6 (i.e. 6, 7, 8, 9, 12) = 15. P = 15/36 = 5/12.
Q3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls.
Solution:
Let number of blue balls = $x$. Total balls = $5 + x$.
P(red) = $5 / (5+x)$. P(blue) = $x / (5+x)$.
Given: P(blue) = 2 $\times$ P(red)
$x / (5+x) = 2 \times [5 / (5+x)]$
Canceling $(5+x)$ from both sides gives $x = 10$.