When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8. a) 5/18 b) 1/18 c) 2/5 d) 1/5

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8. a) 5/18 b) 1/18 c) 2/5 d) 1/5
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c) 2/5
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Description : When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2 

Last Answer : Answer: A) Let the event of getting a greater number on the first die be G. There are 4 ways to get a sum of 9 when two dice are rolled = {(3,6),(4,5),(5,4), (6,3)}. And there are two ways where the number on the ... Now, P(G) = P(G sum equals 9)/P(sum equals 9) = (2/36)/(4/36) = 2/4 =>1/2

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Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

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Last Answer : When two unbiased dice are rolled, the possible out comes are∴ n(S) = 36 Let A : getting a multiple of 2 on one die and a multiple of 3 on the other die. ⇒ A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), ( ... (3, 6), (6, 2), (6, 4)} ⇒ n(A) = 11∴ P(A) = \(rac{n(A)}{n(S)} =rac{11}{36}.\)

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Last Answer : (c) \(rac{11}{36}\)Total number of outcomes when two identical dice are rolled, n(S) = 6 6 = 36 Let A : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die ⇒ A = {(2, 3), (2, 6), (4, 3), (4, ... , 4), (3, 6)} ⇒ n(A) = 11 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{11}{36}\).

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Last Answer : (b) \(rac{5}{24}\)Total number of ways three die can be rolled = 6 6 6 = 216 A larger number than the previous number can be got in the three throws as (1, 2, 3), (1, 2, 4), (1, 2, 5) ( ... , 5, 6). ∴ Total number of favourable cases = 20∴ Required probability =\(rac{20}{216}\) = \(rac{5}{24}\).

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Two trains 240m and 170m in length are running towards each other on parallel tracks. One train has a speed of 51kmph and the other has a speed of 45kmph. In what time after the meet first will they be clear of each other? a) 6.714s b) 12.2 s c) 15.375s d) 16s

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Last Answer : ? ≈ 80 × 35-60 × 85 + 75 × 44 =2800 – 5100 + 3300 = 1000 Answer: b)

119 + 784 ÷ 32 x 116 = ? a) 2961 b) 3121 c) 2564 d) 2861 e) 2941

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