Math, asked by Swarnabhattacharjee, 1 year ago

How many times must a man toss a fair coin, so that the probability of having atleast one head is more than 80%?

Answers

Answered by dainvincible1
1

Let the required coin toses be n

Pr(at least one head in n tosses)>0.8 

                  or

Pr(0 heads in n tosses)=<0.2

Probability of 0 heads in n tosses is (1/2)ⁿ

(1/2)ⁿ =< 0.2 take log on both sides

nlog(0.5)=<log(0.2)

n*(-0.3010)=<(-0.6989) or n=>2.3219

∴ approximately it's 3


maheshwaranlupp0yaxy: Could u explain the first method more clearly
dainvincible1: probability of the least one head in "n' tosses (n= required number of tosses) whi ch greater then 0.8
Answered by pratham107
0
80 times as the probability of getting coins head is not certain.
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