Question

the minimum no.of tosses of a fair coin required, for the probability of getting at least 1 head to be greater than 8/9 ​

Answer 1

Answer:

well the number is 4.

Step-by-step explanation:

To get atleast one head is (2^x)-1/2^x

where x is the number of coins tossed.

thus the answer is 4 where the probability is 15/16.

Answer 2

Given:

A fair coin is tossed a number of times.

To Find:

The minimum no.of tosses of a fair coin required, for the probability of getting at least 1 head to be greater than 8/9 ​

Solution:

Let x be the number of tosses for the probability of getting atleast 1 head to be greater than 8/9 .

• P(at least one head) = 1 - P(No head)
• P(at least one head) ≥ 8/9
• 1 - P( no head) ≥ 8/9
• P( no head ) ≤1/9

We  know ,

• P( no head ) = $P(tails)^{n}$  = $\frac{1}{2}^{n}$  

Therefore,

• minimum value of n so that $\frac{1}{2} ^{n}$ ≤ 1/9
• $2^{n}$ ≥9
• n ≥ 4

The minimum no.of tosses of a fair coin required, for the probability of getting at least 1 head to be greater than 8/9 is 4.