## Book: RD Sharma - Mathematics (Volume 2)

### Chapter: 33. Binomial Distribution

#### Subject: Maths - Class 12th

##### Q. No. 46 of Exercise 33.1

Listen NCERT Audio Books to boost your productivity and retention power by 2X.

46
##### How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Let the man toss the coin n times.

Let p be the probability of getting a head in a toss.

Then, q is the probability of getting a tail in a toss.

Since the coin has only two outcomes, so the probability of getting a head = 1/2

Also, p + q = 1

q = 1 – p

Let X be a random variable that represents a number of occurrence of the head in n tosses of a fair coin.

Then, the probability of getting r number of heads out of total n tosses is given by this Binomial distribution.

P (X = r) = nCrprqn-r

Substituting the value of p and q in the above equation, we get

…(i)

We need to find the number of times the man must toss a fair coin so that the probability of having at least one head is more than 90%.

We can represent it as,

P (getting atleast one head) > 90%

P (X ≥ 1) > 90%

1 – P (X < 1) > 90% [ P (X ≥ 1) = 1 – P (X < 1)]

1 – P (X = 0) > 90% [ P (X < 1) = P (X = 0)]

Put r = 0 in equation (i) and then, substituting it in the above equation.

2n > 10

Now, we need to find the minimum value of n that satisfy this inequality.

Put n = 0.

20 > 10

1 > 10

But 1 10.

Put n = 1.

21 > 10

2 > 10

But 2 10.

Put n = 2.

22 > 10

4 > 10

But 4 10.

Put n = 3.

23 > 10

8 > 10

But 8 10.

Put n = 4.

24 > 10

16 > 10

It is true.

Thus, the minimum n that satisfy this inequality is 4.

Hence, the man should toss the coin 4 or more times.

26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54