Tougher than it looks. Here we go:
Flip a coin. You get H (heads) or T (tails). So two possible outcomes in one flip. If you flip it 5 times, you have 2^5=32 possible outcomes.
How many of these 32 outcomes contain exactly 3 heads?
When we have three heads, we must also have exactly three tails, so your goal is to determine how many combinations of this there are.
Consider one option:
HHHTT
First we need to flip three heads in a row. The odds of this happening are (1/2)^3=1/8
Then we need to flip two tails in a row. Odds: (1/2)^2=1/4. So the probability of getting HHHTT is (1/8)(1/4)=1/32 (good general idea to remember in probability problems: x=and, +=or).
Now consider another option: HHTHT. We flip 2 heads (odds: (1/2)^2=1/4), 1 tails (odds: 1/2), 1 heads (1/2), and another tails (1/2). So the odds of getting this combination are (1/4)(1/2)(1/2)(1/2)=1/32.
The point is, the probability of getting each combination of 3 H's and 2 T's is always the same, 1/32. So, how many possible 5 flip combinations have 3 heads? Well, that would be solved as a combination, 5C3=5!/(3!2!)=10. So the answer is 10(1/32)=10/32=5/16
2007-07-03 20:28:22
·
answer #1
·
answered by Red_Wings_For_Cup 3
·
1⤊
0⤋
Flip a coin. You get H (heads) or T (tails). So two possible outcomes in one flip. If you flip it 5 times, you have 2^5=32 possible outcomes.
How many of these 32 outcomes contain exactly 3 heads?
When we have three heads, we must also have exactly three tails, so your goal is to determine how many combinations of this there are.
Consider one option:
HHHTT
First we need to flip three heads in a row. The odds of this happening are (1/2)^3=1/8
Then we need to flip two tails in a row. Odds: (1/2)^2=1/4. So the probability of getting HHHTT is (1/8)(1/4)=1/32 (good general idea to remember in probability problems: x=and, +=or).
Now consider another option: HHTHT. We flip 2 heads (odds: (1/2)^2=1/4), 1 tails (odds: 1/2), 1 heads (1/2), and another tails (1/2). So the odds of getting this combination are (1/4)(1/2)(1/2)(1/2)=1/32.
The point is, the probability of getting each combination of 3 H's and 2 T's is always the same, 1/32. So, how many possible 5 flip combinations have 3 heads? Well, that would be solved as a combination, 5C3=5!/(3!2!)=10. So the answer is 10(1/32)=10/32=5/16
this is wat my teacher said.
2007-07-03 22:12:16
·
answer #2
·
answered by Anonymous
·
0⤊
0⤋
use the Triangle
to count the successful ways!
Pascal's Triangle to be exact
too much math other ways i says
and prone to making errors
(unless a calculator is used)
the row (6th) that starts with 1,5,10
(5 is the number of coin flips)
there are 6 possible values for the number of Heads
0 to 5 inclusive
read left to right
1 ways = 0 Heads
5 ways = 1 Heads
10 ways = 2 Heads
10 ways = 3 Heads <<< same as exactly 2 Heads
DING DING DING! (5C3 or 5!/3!*2!)
out of 2^5 possible sequences (32)
so the prob = 10/32 = 0.3125
in agreement with others
easy as pie
another triangle!
I (we) wants more pie!
OK
part b)
how about at most 3 Heads?
meaning 3 or less
1+5+10+10=26
26/32 = 0.8125
part c)
how about at least 3 Heads?
meaning 3 or more
10+5+1=16
16/32 = 0.50
Yahoo!
2015-08-13 05:02:30
·
answer #3
·
answered by Anonymous
·
1⤊
0⤋
The probability of getting a head in each toss is 1/2. Similarly the probability of getting a tail is also 1/2.
The probability distribution is binomial.
(H + T)^5 and the coefficient of the term H^3 will be the probability.
2007-07-03 20:22:32
·
answer #4
·
answered by Swamy 7
·
0⤊
0⤋
You have to get three heads and two tails, and there is more than one way of doing this. The formula is
p(3) = 5C3 × 0.5^3 × 0.5^2
where 5C3 = 5!/3!2! = 10 is the number of ways of selecting 3 things out of 5.
This gives the probability of exactly three heads as 0.3125.
.
2007-07-03 20:22:28
·
answer #5
·
answered by rrabbit 4
·
0⤊
0⤋
Number of ways of getting 3 heads in 5 tosses is number of five bit binary words of weight 3 = 10 10/32 = 5/16 TTHHH THTHH THHTH THHHT etc
2016-03-14 22:28:04
·
answer #6
·
answered by Anonymous
·
0⤊
0⤋
The probability is always 50 - 50. Chance confounds the law of averages because each flip is individual, each rotation of the coin will be unlike the previous or the next flip. It is just as likely to produce five "heads" as five "tails". Unless, of course, you know what you're doing. But that's a route that eliminates chance.
2007-07-03 20:27:15
·
answer #7
·
answered by goaltender 4
·
0⤊
0⤋
5C3(1/2)^3(1/2)^2=10/32
2007-07-03 20:18:32
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋