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Question : 13
Total: 29
How many times must a man toss a fair coin, so that the probability of having at least one head is more than 80%?
Solution:
Let the man toss the coin n times. The n tosses are n Bernoulli trials
Probability (p) of getting a head at the toss of a coin is
⇒ p =
⇒ q =
∴ P (X = x) =
C x p n − x q x =
C x (
) n − x (
) x =
C x (
) n
It is given that
P getting at least one head) >
⇒ P (X ≥ 1) > 0.8
⇒ 1 - P (X = 0) > 0.8
⇒ 1 -
C n
> 0.8
⇒
C 0
< 0.2
⇒
< 0.2
⇒2 n >
= 5
⇒2 n > 5 ... (1)
The minimum value of n which satisfies the given inequality is 3.
Thus, the man should toss the coin 3 or more than 3 times.
Probability (p) of getting a head at the toss of a coin is
⇒ p =
∴ P (X = x) =
It is given that
P getting at least one head) >
⇒ P (X ≥ 1) > 0.8
⇒ 1 - P (X = 0) > 0.8
⇒ 1 -
⇒
⇒
⇒
⇒
The minimum value of n which satisfies the given inequality is 3.
Thus, the man should toss the coin 3 or more than 3 times.
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