10 examples of probability with answers
Answers
Answer:
Example 1: A coin is thrown 3 times .what is the probability that atleast one head is obtained?
Sol: Sample space = [HHH, HHT, HTH, THH, TTH, THT, HTT, TTT]
Total number of ways = 2 × 2 × 2 = 8. Fav. Cases = 7
P (A) = 7/8
OR
P (of getting at least one head) = 1 – P (no head)⇒ 1 – (1/8) = 7/8
Example 2: Find the probability of getting a numbered card when a card is drawn from the pack of 52 cards.
Sol: Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36
P (E) = 36/52 = 9/13
Example 3: There are 5 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.
Sol: P (G) × P (R) = (5/12) x (7/11) = 35/132
Example 4: What is the probability of getting a sum of 7 when two dice are thrown?
Sol: Probability math - Total number of ways = 6 × 6 = 36 ways. Favorable cases = (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) --- 6 ways. P (A) = 6/36 = 1/6
Example 5: 1 card is drawn at random from the pack of 52 cards.
(i) Find the Probability that it is an honor card.
(ii) It is a face card.
Sol: (i) honor cards = (A, J, Q, K) 4 cards from each suits = 4 × 4 = 16
P (honor card) = 16/52 = 4/13
(ii) face cards = (J,Q,K) 3 cards from each suit = 3 × 4 = 12 Cards.
P (face Card) = 12/52 = 3/13
Example 6: Two cards are drawn from the pack of 52 cards. Find the probability that both are diamonds or both are kings.
Sol: Total no. of ways = 52C2
Case I: Both are diamonds = 13C2
Case II: Both are kings = 4C2
P (both are diamonds or both are kings) = (13C2 + 4C2 ) / 52C2
Example 7: Three dice are rolled together. What is the probability as getting at least one '4'?
Sol: Total number of ways = 6 × 6 × 6 = 216. Probability of getting number ‘4’ at least one time
= 1 – (Probability of getting no number 4) = 1 – (5/6) x (5/6) x (5/6) = 91/216
Example 8: Find the probability of getting two heads when five coins are tossed.
Sol: Number of ways of getting two heads = 5C2 = 10. Total Number of ways = 25 = 32
P (two heads) = 10/32 = 5/16
Example 9: What is the probability of getting a sum of 22 or more when four dice are thrown?
Sol: Total number of ways = 64 = 1296. Number of ways of getting a sum 22 are 6,6,6,4 = 4! / 3! = 4
6,6,5,5 = 4! / 2!2! = 6. Number of ways of getting a sum 23 is 6,6,6,5 = 4! / 3! = 4.
Number of ways of getting a sum 24 is 6,6,6,6 = 1.
Fav. Number of cases = 4 + 6 + 4 + 1 = 15 ways. P (getting a sum of 22 or more) = 15/1296 = 5/432
Example 10: Two dice are thrown together. What is the probability that the number obtained on one of the dice is multiple of number obtained on the other dice?
Sol:Total number of cases = 62 = 36
Since the number on a die should be multiple of the other, the possibilities are
(1, 1) (2, 2) (3, 3) ------ (6, 6) --- 6 ways
(2, 1) (1, 2) (1, 4) (4, 1) (1, 3) (3, 1) (1, 5) (5, 1) (6, 1) (1, 6) --- 10 ways
(2, 4) (4, 2) (2, 6) (6, 2) (3, 6) (6, 3) -- 6 ways
Favorable cases are = 6 + 10 + 6 = 22. So, P (A) = 22/36 = 11/18
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