A fair coin is tossed 10 times. Ture of False, and explain:
1) The chance of getting 10 heads in a row is 1/1,024
2)Given that the first 9 tosses were heads, the chances of getting 10 heads in a row is 1/2.
I said that they were both True because...
1) (1/2)^10 = 1/1024 ....but maybe it's false...because coin tosses are ALWAYS (?) independent? so it could be just 1/2? help?
2) because the 10th toss is independent of the previous 9 tosses?? is this the same question or no???
thanks for the help in advance.
2007-04-01 05:49:53 · 7 answers · asked by Captain Whiskerboy Litterbox 3 in Science & Mathematics ➔ Mathematics
1) You were right when you said true. Think about it: the probability of getting heads when you toss one coin will always be 1/2... but getting heads TEN times in a row? That would almost never happen! The truth is that events HAVE to be independent for you to be able to multiply the probabilities like that. It's 1/2 likely to get heads the first time, but 1/4 likely to get heads both times, and 1/8 likely to get heads all three times...
2) True, because the events are independent. This means that one coin toss does NOT depend on the outcome of the toss before it. However, if you get 9 heads in a row, chances are the stupid coin's rigged.
2007-04-01 05:56:34 · answer #1 · answered by dac2chari 3 · 0⤊ 0⤋
Both are true and I will give you credit for the second answer but not first since you were not confident. The probability of getting a head is 1/2 for each trial with a fair coin. So if we toss it ten times, the results follow the binomial distibution, (pH + pT)^10 where pH is the probability of heads each time and is 1/2. So, the probability for all 10 heads to turn up is indeed, 1/2 raised to the power of 10, or 1/1024.
When all 9 times heads came up, that was an event with a probability of 1/512 and the next trial again has a probability of 1/2.
2007-04-01 05:57:24 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
the two are real and that i visit grant you credit for the 2nd answer yet no longer first via fact which you weren't helpful. The risk of having a head is a million/2 for each trial with a honest coin. So if we toss it ten circumstances, the outcomes follow the binomial distibution, (pH + pT)^10 the place pH is the risk of heads each and every time and is a million/2. So, the risk for all 10 heads to take place is unquestionably, a million/2 raised to the means of 10, or a million/1024. whilst all 9 circumstances heads got here up, that replaced into an experience with a risk of a million/512 and the subsequent trial returned has a risk of a million/2.
2016-11-25 19:27:55 · answer #3 · answered by cornelius 4 · 0⤊ 0⤋
1. True. (1/2)^10 = 1/1024
2. True. Coin tosses are independent events.
Even though coin tosses are independent events, the probability of an event that involves multiple coin tosses will be the result of multiplying their individual probabilities, so your hunch was correct on #1.
2007-04-01 05:52:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋
1st is correct....1/2^10 = 1/1024
2nd is definitely true....no matter which toss you are at, the chances of heads is always 1/2
2007-04-01 05:56:58 · answer #5 · answered by Anonymous · 0⤊ 0⤋
yes, u are most certainly correct, i am in an ap stats class and we have done problems like that. each is independent so it would be 1/2 to the 10th, and since the 10th flip is independent, its is still .5 chance...
2007-04-01 05:55:04 · answer #6 · answered by Chris H 2 · 0⤊ 0⤋
Those both look pretty good to me.
2007-04-01 05:54:13 · answer #7 · answered by I-Eat-Paint-Chips 3 · 0⤊ 0⤋