If we toss four fair coins, what is the probability that we get more heads than tails?
HHHH TTTT
HHTT TTHH
HTHT THTH
HHHT TTTH
HTTH THHT
Would the answer be 2/10?
I am a little confused.
Thanks for the help!
With 4 coins, more heads than tails
---> 3H1T or 4H
prob 3 of 4 heads = C(4,3)(1/2)^3 (1/2)
= 4(1/8)(1/2) = 1/4
prob 4 out of 4 heads = (1/2)^4 = 1/16
prob(your event) = 1/4 + 1/16 = 5/16
aside:
prob(no heads) = (1/2)^4 = 1/16
prob(1 head) = C(4,1)(1/2) (1/2)^3 = 1/4
prob(2 heads) = C(4,2)(1/2^2 (1/2)^2 = 6(1/16)
= 3/8
prob(3 heads) = 1/4 , see above
prob(4 heads) = 1/16
note: 1/16 + 1/4 + 3/8 + 1/4 + 1/16 = 1
easier grid for me
h -> 0 1 2 3 4
t -> 4 3 2 1 0
I see five possible outcomes of which 2 work so I get 2/5
HHTT is the same here as TTHH or HTTH
I missed that this is a binomial distribution problem. The combinations are not equally likely.
i just did this problem and i got 5/16, it doesnt matter if its fore heads or tails the answer is this if you are throwing 4 coins yw.
To find the probability of getting more heads than tails when tossing four fair coins, you can list all the possible outcomes and count how many of them have more heads than tails.
You correctly listed all the possible outcomes of tossing four coins. Now let's identify the outcomes where there are more heads than tails:
HHHH (4 heads, 0 tails)
HHTT (2 heads, 2 tails)
HTHT (2 heads, 2 tails)
HHHT (3 heads, 1 tail)
HTTH (2 heads, 2 tails)
Out of the 10 possible outcomes, there are 5 outcomes where there are more heads than tails. So, the probability of getting more heads than tails is 5/10, which can be simplified to 1/2 or 50%.
Therefore, your answer of 2/10 is not correct. The correct probability is 1/2 or 50%.