Q. 12

Consider the expe

Answer :

Given: A coin is tossed: If the coin shows tail, it is tossed again. If it shows head, then a die is thrown.



Hence, different value of probabilities are



We need to find the probability that the die shows a number greater than 3, given that there is atleast one head.


Now, F: Number greater than 3 on the die


E: atleast one head


We need to find P(E|F)


E = {(H, 4), (H, 5), (H, 6)}


P(E) = P(H, 4) + P(H, 5) + P(H, 6)





F = {(T, H), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}


P(F) = P(T, H) + P(H, 1) + P(H, 2) + P(H, 3) + P(H, 4) + P(H, 5)


+ P(H, 6)







Also, EF = Number greater than 3 and also atleast one head


EF = (H, 4), (H, 5), (H, 6)


So, P(EF) = P(H, 4) + P(H, 5) + P(H, 6)





Now, we know that the Property of Conditional Probability:


P(E|F) × P(F) = P(E F)






OR


Let X: number of heads appearing


Coin toss is a Bernoulli trial


So, X has a binomial distribution


P(X = x) = nCx qn–x px


Here,


n = number of coin tosses


p = probability of head



q = 1 – p




Substituting the values, we get



We know that, when the base is same the powers get add




Now, we have to find that how many times must a man toss a fair coin so that the probability of having atleast one head is more than 90%


So, given P(X ≥ 1) > 90%


We need to find n


Now,


P(X ≥ 1) > 90%


1 – P(X = 0) > 90%


[Here, x = 0]







2n < 10


We know that 24 = 16 < 10


21 = 2


22 = 2 × 2 = 4


23 = 2 × 2 × 2 = 8


24 = 2 × 2 × 2 × 2 = 16


So, n ≥ 4


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