You can read off the outcomes from Pascal's Triangle. For 5 tosses consult the fifth row
........1 1
.......1 2 1
......1 3 3 1
.....1 4 6 4 1
....1 5 10 10 5 1
There is
1 chance in 32 of 5 heads and 0 tails
5 chances in 32 of 4 heads and 1 tail
10 chances in 32 of 3 heads and 2 tails
10 chances in 32 of 2 heads and 3 tails
5 chances in 32 of 1 head and 4 tails
1 chance in 32 of 0 heads and 5 tails
2006-11-19 12:13:54
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answer #1
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answered by Anonymous
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To calculate, multiply the probabilities for each flip, five total times.
1/2 x 1/2 x 1/2 x 1/2 x 1/2
or (1/2)^5
Yes, it's 1/32.
2006-11-19 11:59:26
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answer #2
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answered by mr_mumbles_nyc 3
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Yes. Each "event" is independent, so to get the probablity of ALL occuring, you just multiply them, so in this case 1/2 * 1/2......
= 1/(2^5) = 1/32
2006-11-19 12:06:44
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answer #3
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answered by clueless_nerd 5
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You are correct - ignoring the extremely small but real possibility of the coin actually landing on its side. I have seen it happen once. A real jaw dropper.
once 1 in 2
twice 1 in 4
thrice 1 in 8
four times 1 in 16
five times 1 in 32
2006-11-19 12:01:44
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answer #4
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answered by Mark T 6
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Yup! (1/2)^5=1/32
2006-11-19 13:02:15
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answer #5
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answered by yupchagee 7
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you have 6 possible outcomes 0 heads 5tails, 1h 4t, 2h 3t, 3h 2t, 4h 1t, 5h 0t. when you consider that heads occurs two times in 4 possiblilties yet precisely two times in a single the probability is a million in 4.
2016-10-04 03:47:44
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answer #6
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answered by ? 4
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yes it is becouse these evnts are statistically independent
2006-11-19 12:01:49
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answer #7
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answered by an ugly mind 2
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