If 3 coins are tossed, what is the probability of getting a head and 2 tails?

Answer 1

i guess its 3/8
totally there are 2 x 2 x 2 = 8 possibities
The probabilities with 1 head and 2 tails are HTT, THT and TTH
Therefore the answer is 3/8

Answer 2

There are a lot of bad answers here--I'm not sure you are going to be able to pick out the correct one2. When you toss k coins (in this case k=3) there are 2^k = 2*2*...*2 different ways for the k coins to land, because each coin independently has two choices of how to land, either H or T, so a sequence of k choices is made. Thus there are 8 ways for the coins to land. If you list all the 8 outcomes for coins 1,2,3 in order, left to right--

hhh
thh
hth
hht
.
.
.
you will see that exactly 3 of the 8 have two t's. Since all 8 ways of landing are equally likely, the probability is 3/8.

This answer is correct.

Answer 3

3/8

There are 8 ways that three coins tosses can come out. 3 of them have 1 head and 2 tails.

Answer 4

Donaldgirod is exactly correct.

Getting one H and two T when tossing three coins has a probability of:

3C1 (three coins; one H) * (1/2) (odds of H) * (1/2)^2 (odds of two independent T) =

3 * (1/2) * (1/4) = 3/8.

Or you can write down all 8 possibilites and count every instance of one H and two T.

Answer 5

There are 8 possiblities, each and every the two probable: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT 3 of those have precisely 2 tails: 3/8. 4 of those have 2 or greater tails: a million/2. that's as much as you to choose for what the question is looking.

Answer 6

1 out of 3

Answer 7

1/8
1/2*1/2*1/2
because there is a one out of two chance for each coin toss and they are not dependent events.

Answer 8

The sample space for three coin toss will be -
S = {TTH, THT, HTT, TTT, HHH, THH, HTH, HHT}

=> n(S) = 8 and n(2T) = 3

So, probability of getting a head and 2 tails is -

P[2T] = n(2T)/n(S) = 3/8 = 0.375

Answer 9

1/8 multiply posiple out comes by how many times. (x=outcomes times number of flips. 1/x=probability.)

Answer 10

1/4 i think