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, E ∩ F = Number greater than 3 and also atleast one head

E ∩ F = (H, 4), (H, 5), (H, 6)

So, P(E ∩ F) = 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) = ^{n}C_{x} q^{n–x} p^{x}

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]

⇒ 2^{n} < 10

We know that 2^{4} = 16 < 10

2^{1} = 2

2^{2} = 2 × 2 = 4

2^{3} = 2 × 2 × 2 = 8

2^{4} = 2 × 2 × 2 × 2 = 16

So, n ≥ 4

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