

InterviewSolution
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A die is thrown. Find the probability of getting:(i) a prime number(ii) 2 or 4(iii) a multiple of 2 or 3(iv) an even prime number(v) a number greater than 5(vi) a number lying between 2 and 6 |
Answer» Given: A dice is thrown once Required to find: (i) Probability of getting a prime number (ii) Probability of getting 2 or 4 (iii) Probability of getting a multiple of 2 or 3. (iv) Probability of getting an even number (v) Probability of getting a number greater than five. (vi) Probability of lying between 2 and 6 Total number on a dice is 6 i.e., 1, 2, 3, 4, 5 and 6. (i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3. We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, probability of getting a prime number = \(\frac{3}{6}\) = \(\frac{1}{2}\) (ii) For getting 2 and 4, clearly the number of favourable outcomes is 2. We know that Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting 2 or 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\) (iii) Multiple of 2 are 3 are 2, 3, 4 and 6. So, the number of favourable outcomes is 4 We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting an multiple of 2 or 3 = \(\frac{4}{6}\) = \(\frac{2}{3}\) (iv) An even prime number is 2 only. So, the number of favourable outcomes is 1. We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting an even prime number = \(\frac{1}{6}\) (v) A number greater than 5 is 6 only. So, the number of favourable outcomes is 1. We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting a number greater than 5 = \(\frac{1}{6}\) (vi) Total number on a dice is 6. Numbers lying between 2 and 6 are 3, 4 and 5 So, the total number of numbers lying between 2 and 6 is 3. We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting a number lying between 2 and 6 = \(\frac{3}{6}\) = \(\frac{1}{2}\) |
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