Ex 13.4, 5 - Find probability distribution of number of successes

Ex 13.4, 5 - Chapter 13 Class 12 Probability - Part 2
Ex 13.4, 5 - Chapter 13 Class 12 Probability - Part 3 Ex 13.4, 5 - Chapter 13 Class 12 Probability - Part 4 Ex 13.4, 5 - Chapter 13 Class 12 Probability - Part 5

  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

Ex 13.4, 5 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 X = 1 Ex 13.4, 5 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 dieSince 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) = 11/36 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 – 11/36 = 25/36 ∴ P(X = 0) = 25/36 So, our probability distribution is

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