A math problem thats really buggin me
Three fair coins are tossed, what is the probability that there are at least 2 heads. Wouldn't it be 1:1? Seeing as how the sample space is this
HTT, HHT, HHH, HTH, THH,TTH, THT, TTT
thanks in advance...
2006-11-15 20:48:24 · 10 answers · asked by jakesly 1 in Science & Mathematics ➔ Mathematics
sample space=8
number of outcomes with at
least 2 heads=4
P(at least 2 heads)=4/8=1/2
odds:even money
i hope that this helps
2006-11-15 20:57:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
To get >= 2 heads, you need:
1) 2 heads (& 1 tail), plus
2) 3 heads.
Probability of 2 heads & 1 tail = 1/2 x 1/2 x 1/2 x 3 (can come in different combinations of HHT, HTH, or THH) = 3/8 plus
probability of 3 heads = 1/2 x 1/2 x 1/2 = 1/8
Therefore probability of at least 2 heads = 3/8 + 1/8 = 1/2.
2006-11-15 21:08:22 · answer #2 · answered by dreamofyz 2 · 0⤊ 0⤋
can't be 1:1, that means it is always going to happen
you said at least 2 which means you can have 3, so we will include the set that has all heads as well as the sets with just 2, so by looking at what you have we have 4 sets that have 2 or more heads, we have a total of 8 possibilities, so the probability of have 2 or more heads face up is 4:8 or it you want a percentage, divide 4 by 8 and you get .50 move the decimal 2 places to the LEFT and put a percent sign at the end, 50%
2006-11-15 21:06:38 · answer #3 · answered by brandonjb2004 2 · 0⤊ 0⤋
htt hth hht hhh ttt tth tht thh
3 two heads
1 three heads
total 4 from 8 = 50%
2006-11-15 21:21:26 · answer #4 · answered by abo balal 1 · 0⤊ 0⤋
the probability:
HHH*
HHT*
HTH*
THH*
HTT
TTH
THT
TTT
there are 8 possibilities in total.
there are 4 different combination of getting at least 2 H
so the chance will be = 4:8 = 50%
2006-11-15 22:32:34 · answer #5 · answered by fii 3 · 0⤊ 0⤋
No, if you are talking about probability, the denominator of your answer should be total no of possible combinations
Answer in this case will be 4/8 = 1/2 = 0.5
2006-11-15 21:03:41 · answer #6 · answered by yogen p 2 · 0⤊ 0⤋
no its atleast 2 heads means 2 and more than 2 heads.
which means
HHT,HHH,HTH,THH
ie 4/8=.5
2006-11-15 21:36:56 · answer #7 · answered by krishna 4 · 0⤊ 0⤋
SAMPLE SPACE
HHH,HHT,HTT,TTT,TTH,THH,THT,HTH
TOTAL OUTCOMES=8
FAVOURABLE OUTCOMES=4
HENCE, P(HH)=4/8=1/2
2006-11-15 21:08:24 · answer #8 · answered by wizkid!!! 3 · 0⤊ 0⤋
50%.
I was about to check mathematically when I realised that with fair coins it is exactly the same as the probability of two or more tails. Therefore symmetry takes you straight to 50%
2006-11-15 21:06:48 · answer #9 · answered by Pedestal 42 7 · 0⤊ 0⤋
yes you are sort of correct
Probability = 1/2
Odds are 1:1
2006-11-15 20:52:15 · answer #10 · answered by Wal C 6 · 0⤊ 2⤋