Bayes theorem

Chapter 13 Class 12 Probability (Term 2)
Concept wise

### Transcript

Ex 13.3, 10 Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?Let A : 1, 2, 3, 4 appear on the die B : 5, 6 appear on the die C : exactly one head is obtained We need to find the Probability that if exactly one head is obtained on the toss of a coin, she threw 1, 2, 3 or 4 with the die i.e. P(A|C) P(A|C) = (π(π΄) ." " π(πΆ|π΄))/(π(π΄) . π(πΆ|π΄) + π(π΅) . π(πΆ|π΅) ) "P(A)" = Probability that 1, 2, 3 or 4 appear on the die = 4/6 = π/π "P(C|A)" = Probability that exactly one head is obtained, if 1, 2, 3 or 4 appear on the die = π/π "P(B)" = Probability that 5 or 6 appear on the die = 2/6 = π/π "P(C|B)" = Probability that exactly one head is obtained, if 5 or 6 appear on the die = π/π Puttinag values in formula, P(A"|"C) = (2/3 Γ 1/2)/( 2/3 Γ 1/2 + 1/3 Γ 3/8 ) = (1/3 Γ 1/2 Γ 2)/( 1/3 Γ 1/2 [2 + 3/4 ] ) = 2/( 2 + 3/4 ) = 2/(11/4) = π/ππ Therefore, required probability is 8/11

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#### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.