Question

20. A coin is tossed 10 times. Find the
probability of getting only one tail .​

Answer 1

Hope this helps

Probability= 1 - probability of 0 tails (or 10 heads) in 10 tosses

= 1- (1/2)*(1/2)*(1/2)* ten times

=1 - (1/2)^10

=1023/1024.

Explanation:

This is a Binomial Distribution .

Probability of r success from n number of trials is given by

P(n,r)=nCr∗pr∗(1−p)n−r

where p is probability that the event will occur.

Probability that tail will occur

p=12

When p=1/2 then probability can be given by:

P(n,r)=nCr∗pn

n=10, number of tosses

As per question, the probability that number of tails would be between 1 to 10 will be

P(10,1)+P(10,2)+P(10,3)+...+P(10,10)

But,

Total Probability=1

P(10,0)+ { P(10,1)+P(10,2)+P(10,3)+...+P(10,10) } = 1

P(10,1)+P(10,2)+P(10,3)+...+P(10,10)=1-P(10,0)

nCr=n!r!∗(n−r)!

Probability=1-P(10,0)

P=1−10C0/210

P=1−11024

P=10231024

Answer 2

Answer:

Step-by-step explanation:

0.01 tail