# Ex 13.4, 5

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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Ex 13.4, 5 (Method 1) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 Tossing a Die is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 Here, n = number of tosses of die = 2 p = Probability number is greater than 4 = 26 = 13 q = 1 – p = 1 – 13 = 23 Hence, ⇒ P(X = x) = 2Cx 𝟏𝟑𝒙 𝟐𝟑𝟐 − 𝒙 Let X : Number of number greater than 4 appears on die When we throw two dies, there can be three cases, • both number greater than 4 • 1 number greater than 4 • no number greater than 4 So, values of X can be 0, 1, 2 Putting values in (1) P(X = 0) = 2C0 130 232 − 0 = 1 × 1 × 232 = 49 P(X = 1) = 2C1 131 232 − 1 = 2 × 13 × 23 = 49 P(X = 2) = 2C2 132 232 − 2 = 1 × 19 × 1 = 19 So, the probability distribution is Ex 13.4, 5 (Method 2) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 Let X : Number of number greater than 4 appears on die When we throw two dies, there can be three cases, • both number greater than 4 • 1 number greater than 4 • no number greater than 4 So, values of X can be 0, 1, 2 S = (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) So, our probability distribution is Ex 13.4, 5 (Method 1) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (ii) six appears on at least one die Let X : Number of times six appears on atleast one die Probability of getting 6 = 16 Probability of not getting 6 = 1 – 16 = 56 P(X = 0) i.e. Probability that in both die, we do not get 6 P(X = 0) = P(not getting 6) × P(not getting 6) P(X = 0) = 56 × 56 = 2536 P(X = 1) i.e. Probability that we get 6 in atleast one die P(X = 1) = P(we get 6 in atleast one die) = 1 – P(we do not get 6) = 1 – P(X = 0) = 1 – 2536 = 1136 So, probability distribution is Ex 13.4, 5 (Method 2) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (ii) six appears on at least one die Since pair of die are thrown, There can be two cases, • Six does not appear at all • Six appears on atleast one die Hence, X = 1 means six appears on atleast one die X = 0 means six does not appear Finding P(X = 1) i.e. Probability that six appears on atleast one die S = (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) P(X = 1) = 1136 Finding P(X = 0) i.e. Probability that six does not appear P(X = 0) = Probability that six does not appear = 1 – Probability that six appears atleast once = 1 – P(X = 1) = 1 – 1136 = 2536 ∴ P(X = 0) = 2536 So, our probability distribution is

Chapter 13 Class 12 Probability

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.