Question

a fair coin is tossed 8 times. Find the probability of getting no head and at least one head​

Answer 1

Given :

• a fair coin is tossed 8 times

To find :

the probability of getting

1. no head
2. at least one head

Solution :

a fair coin is tossed thrice

so, simple space

= 2³

= 8

I. e. (HHH) ,( HHT), (HTH) ,(THH) , (HTT), (THT), (TTH), (TTT)

1) No head

= (TTT)

= 1

probability = ⅛

2) at least one head

= (HHH) ,( HHT), (HTH) ,(THH) , (HTT), (THT), (TTH)

= 7

Probability = 7/8

Answer 2

Step-by-step explanation:

Given :-

• A fair coin is tossed 8 times .

Find to :-

Find the probability of getting -

• no head
• at least one head

Solution :-

• tossed = 8 times
• $8 \:=\: {2}^{3}$

( 2 × 2 × 2 = 8 )

1 = (HHH)

2 = (HHT)

3 = (HTH)

4 = (THH)

5 = (HHT)

6 = (THT)

7 = (TTH)

8 = (TTT)

1) no head

(TTT)

So, your answer is 1

Hence,

Probability = $\frac{1}{8}$

2) at least one head

(HHH) (HHT) (HTH) (THH) (HHT) (THT) (TTH)

So, your answer is 7

Hence,

Probability = $\frac{7}{8}$