You toss an unfair coin, where the probability of tails is 0.8. find the smallest number n of times you have to toss the coin so that the probability at least one head goes over 50%.

Guest Mar 19, 2020

#2**+1 **

If you don't get at least one head, you must get all tails.

To find the probability of getting at least one head, subtract the probability of getting all tails from 1.000.

One toss: 1.000 - all heads = 1.000 - 0.80 = 0.20

Two tosses: 1.000 - all heads = 1.000 - 0.80x0.80 = 0.36

Three tosses: 1.000 - all heads = 1.000 - 0.80x0.80x 0.80 = 0.488

Four tosses: 1.000 - all heads = 1.000 - 0.80x0.80x.80x0.80 = 0.5904

It takes four tosses.

geno3141 Mar 19, 2020

#1**0 **

if tails prob=.8 then heads =.2

2*n = 5+

n is greater then 2.5

but you cant toss a coin .5 times

so **n=3 tosses**

i think this is right i could be wrong

Guest Mar 19, 2020

#2**+1 **

Best Answer

If you don't get at least one head, you must get all tails.

To find the probability of getting at least one head, subtract the probability of getting all tails from 1.000.

One toss: 1.000 - all heads = 1.000 - 0.80 = 0.20

Two tosses: 1.000 - all heads = 1.000 - 0.80x0.80 = 0.36

Three tosses: 1.000 - all heads = 1.000 - 0.80x0.80x 0.80 = 0.488

Four tosses: 1.000 - all heads = 1.000 - 0.80x0.80x.80x0.80 = 0.5904

It takes four tosses.

geno3141 Mar 19, 2020