Alice rolls two dice. What is the probability of obtaining at least one 6?Select one:Select the correct option.a. 1/24b. 10/36c. 1/6d. 11/36 Solution: option (1) 1/9 If two dice are thrown simultaneously, the total number of sample space is 36 Favourable outcomes = (1, 4), (4, 1), (2, 3) and (3, 2) Therefore, the required probability = 4/ 36 = 1/9.
`1/6` Explanation: When two dice are thrown simultaneously, all possible outcomes are: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)(5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) Number of all possible outcomes = 36 Let E be the event of getting a doublet. Then the favourable outcomes are: (1,1), (2,2) , (3,3) (4,4), (5,5), (6,6) Number of favourable outcomes = 6 ∴ P(getting a doublet) = P ( E) = `6/36 = 1/6` Page 2`3/4` Explanation: When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT. Total number of possible outcomes = 4 Let E be the event of getting at most one head. Then, the favourable outcomes are HT, TH and TT. Number of favourable outcomes = 3 ∴ P(getting at most 1head) = `("Number of favourable outcomes")/"Number of all possible outcomes" = 3/4` Page 3`3/8` Explanation: When 3 coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, THT, HTT, TTH and TTT. Total number of possible outcomes = 8 Let E be the event of getting exactly two heads. Then, the favourable outcomes are HHT, THH, and HTH. Number of favourable outcomes = 3 ∴ Probability of getting exactly 2 heads = P(E) =`3/8` Select the correct option from the given alternatives : Two dice are thrown simultaneously. Then the probability of getting two numbers whose product is even is `3/4` Explanation; Two dice are thrown. ∴ n(S) = 36. Getting two numbers whose product is even, i.e., one of the two numbers must be even. Let event A: Getting even number on first dice. event B: Getting even number on second dice. ∴ n(A) = 18, n(B) = 18, n(A ∩ B) = 9 ∴ Required probability = P(A ∩ B) = `("n"("A") + "n"("B") - "n"("A" ∩ "B"))/("n"("S"))` = `(18 + 18 - 9)/36` = `27/36` = `3/4` Concept: Concept of Probability Is there an error in this question or solution?
Correct Answer: D Solution :
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Correct Answer: C Solution : Number x can be selected in three ways and corresponding to each such way there are three ways of selecting number y. Therefore, two numbers can be selected in 9 ways as listed below (1,1), (1,4), (1,9), (2,1), (2,4), (2,9), (3,1), (3,4), (3,9) So, total numbers of possible outcomes = 9 The product xy will be less than 9, if x and y are chosen in one of the following ways (1, 1), (1,4), (2,1), (2,4), (3.1) \[\therefore \]Favourable number of elementary events = 5 Hence, required probability \[=\frac{5}{9}\]Page 16
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Correct Answer: B Solution : [A] Probability of getting number 5 in throwing a dice \[=\frac{1}{6}\] [B] Probability of getting three heads in a single throw of coin = 0 [C] Probability of getting the sum of the number as 7 [(3, 4), (4, 3), (1, 6), (6, 1), (2, 5), (5, 2)] when two dice are thrown \[=\frac{6}{36}\] [D] Probability of occurence of two sure independent events is \[\frac{2}{7}\].Page 22
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