Ex 13.1, 15 - Chapter 13 Class 12 Probability
Last updated at April 16, 2024 by Teachoo
Ex 13.1
Ex 13.1, 2
Ex 13.1, 3 Important
Ex 13.1, 4
Ex 13.1, 5
Ex 13.1, 6 (i)
Ex 13.1, 6 (ii) Important
Ex 13.1, 6 (iii)
Ex 13.1, 7 (i)
Ex 13.1, 7 (ii)
Ex 13.1, 8
Ex 13.1, 9
Ex 13.1, 10 (a) Important
Ex 13.1, 10 (b) Important
Ex 13.1, 11
Ex 13.1, 12 Important
Ex 13.1, 13 Important
Ex 13.1, 14
Ex 13.1, 15 You are here
Ex 13.1, 16 (MCQ) Important
Ex 13.1, 17 (MCQ) Important
Last updated at April 16, 2024 by Teachoo
Ex 13.1, 15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.A die is thrown If multiple of 3 comes up, the die is thrown again If any other number comes up, a coin is tossed Thus, our event looks like Die is thrown We need to find the Probability of the coin showing a tail, given that at least one die shows a 3. E : Coin shows a tail F : At least one die shows 3 We need to find P(E|F) We need to find the Probability of the coin showing a tail, given that at least one die shows a 3. E : Coin shows a tail F : At least one die shows 3 We need to find P(E|F) Event E E = { (1, T), (2, T), (4, T), (5, T)} P(E) = P(1, T) + P(2, T) + P(4, T) + P(5, T) = 1/12+1/12+1/12+1/12 = 4/12 = 𝟏/𝟑 Event F F = { (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)} P(F) = P(3, 1) + P(3, 2) + ... + P(3,6) + P(6, 3) = 1/36+1/36+1/36+1/36+1/36+1/36+1/36 = 𝟕/𝟑𝟔 Also, "E"∩"F" = 𝝓 So, P("E" ∩"F" )=0 Now, P(E|F) = 𝑃(𝐸 ∩ 𝐹)/(𝑃(𝐹)) = 0/(7/36) = 0