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Ex 15.1, 13 - Chapter 15 Class 10 Probability (Term 1)

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Ex 15.1, 13 - A die is thrown once. Find probability - Die

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Ex 15.1, 13 - Chapter 15 Class 10 Probability - Part 2

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Ex 15.1, 13 - Chapter 15 Class 10 Probability - Part 3

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Transcript

Ex15.1, 13 A die is thrown once. Find the probability of getting (i) a prime number; Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Number of possible outcomes of a dice = 6 Prime number is a number not divisible by any number except itself Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice = 3 Probability of getting a prime number = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘  π‘€β„Žπ‘’π‘Ÿπ‘’ π‘π‘Ÿπ‘–π‘šπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘π‘œπ‘šπ‘’π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ ) = 3/6 = 1/2 Ex15.1, 13 A die is thrown once. Find the probability of getting (ii) a number lying between 2 and 6; Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Number of possible outcomes of a dice = 6 Numbers lying between 2 and 6 = 3, 4, 5 Total numbers lying between 2 and 6 = 3 Probability of getting a number lying between 2 & 6 = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘  π‘€β„Žπ‘’π‘Ÿπ‘’ π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 2 & 6)/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ ) = 3/6 = 1/2 Ex15.1, 13 A die is thrown once. Find the probability of getting (iii) an odd number. Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Number of possible outcomes of a dice = 6 Numbers which are odd = 1,3, 5 Total numbers which are odd = 3 Probability of getting an odd number = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘  π‘€β„Žπ‘’π‘Ÿπ‘’ π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑖𝑠 π‘Žπ‘› π‘œπ‘‘π‘‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿ)/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ ) = 3/6 = 1/2

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