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7 answers

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 · answer #1 · answered by Anonymous · 0 1

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 · answer #2 · answered by mr_mumbles_nyc 3 · 0 1

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 · answer #3 · answered by clueless_nerd 5 · 0 0

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 · answer #4 · answered by Mark T 6 · 1 1

Yup! (1/2)^5=1/32

2006-11-19 13:02:15 · answer #5 · answered by yupchagee 7 · 0 1

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 · answer #6 · answered by ? 4 · 0 0

yes it is becouse these evnts are statistically independent

2006-11-19 12:01:49 · answer #7 · answered by an ugly mind 2 · 1 1

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