Solution: We use the basic formula of probability to solve the problem. Number of outcomes on throwing a die is (1, 2, 3, 4, 5, 6) = 6 Number of prime numbers on dice are 1, 3 and 5 = 3 (i) Probability of getting a prime number = Number of prime numbers/total number of outcomes = 3/6 = 1/2 (ii) Numbers lying between 2 and 6 are 3, 4, 5 = 3 Probability of getting a number lying between 2 and 6 = Number lying between 2 and 6/total number of outcomes = 3/6 = 1/2 (iii) Total number of odd numbers are 1, 3 and 5 = 3 Probability of getting a odd number = Number of odd numbers/total number of outcomes = 3/6 = 1/2 ☛ Check: NCERT Solutions Class 10 Maths Chapter 15 Video Solution: 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.NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 Question 13 Summary: If a die is thrown once, then the probability of getting (i) a prime number, (ii) a number lying between 2 and 6, and (iii) an odd number are 1/2, 1/2, and 1/2 respectively. ☛ Related Questions: Math worksheets and Answer Hint: In this question it is given that if a die rolled once, then we have to find the probability of getting an odd prime number. So for this we need to know the expression of probability, which is,Probability(P)=$$\dfrac{\text{Number of favourable outcomes} }{\text{Total number of outcomes} }$$.....(1)So from the given data we have to find the all possible outcomes and the number of possible outcomes. Complete step-by-step solution: As we know that a die is a six-sided cube with the numbers 1-6 placed on the faces.So from here we can say that if we throw a die then among these six faces one face must occur. Therefore, Total number of possible outcomes = 6Now the favourable outcome is to get an odd prime number, so between 1 to 6 the odd primes are 3 and 5, i.e, only two.So the number of favourable outcomes =2Therefore, we can say that,$$\text{Probability} \left( P\right) =\dfrac{\text{Number of favourable outcomes} }{\text{Total number of possible outcomes} }$$$$=\dfrac{2}{6}$$$$=\dfrac{1}{3}$$So we can say that the probability of getting an odd prime number is $$\dfrac{1}{3}$$.Note: To solve this type of question you need to know that in mathematics, prime numbers are whole numbers which are greater than 1, that have only two factors, 1 and the number itself and they are divisible only by the number 1 or itself. For example: 2, 3, 5, 7 and 11 are the first few prime numbers.Also you need to have the basic idea about odd numbers, so odd numbers are those numbers which are not divisible by 2 or we can say that when divided by 2, leave a remainder 1.For example: 1, 3, 5, 7, …...No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 0 |
What is the probability of getting an odd prime number when a dice is thrown?
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