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Question about probability

T

tontod

Diamond Member
I know how to calculate probability with a fair coin, i.e. probability of either head or tail is 0.5. But, I have to calculate the probability that I will get 1 head AND 9 tails by flipping an UNFAIR coin 10 times with probability of head p =0.1 and tail q = 0.9.

No flames please 😛
 
Originally posted by: amdfanboy
.1 *.9 *.9 *.9 ..... ?
Click to expand...
That's the probability that he gets heads on the first flip and tails the next nine....

you want the sum of the probability of all ten permutations...so basically 10 times what you wrote

P(H-T-T-T-T-T-T-T-T-T-T) + P(T-H-T-T-T-T-T-T-T-T-T) + P(T-T-H-T-T-T-T-T-T-T-T) + ... + P(T-T-T-T-T-T-T-T-T-T-H)
=(0.1 x 0.9^9) + (0.9 x 0.1 x 0.9^8) + (0.9^2 x 0.1 x 0.9^7) + ... + (0.9^9 x 0.1)
= 10 (0.1 x 0.9^9)
=0.9^9
=pull out your calculator
 
Originally posted by: alexjohnson16
Are you talking like a weighted coin?

I don't catch you.
Click to expand...
This coin doesn't exist in real life...or would at least take some serious engineering to get exactly right...
 
to quote from 2 stupid dogs... "awww, isn't that cute... BUT IT"S WRONG!"

.1 * .9 * .9 *.9 *. . . is the probability of getting a heads on the first toss and tails on the rest of the tosses.
Since you didn't specify that it had to occur in that order, it could be 4 tails, heads, followed by 5 tails,
in which case, the probability of that happening would be .9*.9*.9*.9*.1*.9*.9*.9*.9*.9*

And, then there are other ways. In fact, there are 10 different ways to rearrange it, therefore, you're going to have to multiply the presumed answer above by 10.


Note, if it was 2 heads, 8 tails, there would be 10combination2 ways to arrange it, .1^2 * .9^8

edit: the general formula for these types of experiments is

nCr * (probability of desired outcome)^r * (1-probability of desired outcome)^(n-r)

Where you want the desired outcome to occur exactly r times in n trials.

 
Originally posted by: b0mbrman
Originally posted by: alexjohnson16
Are you talking like a weighted coin?

I don't catch you.
Click to expand...
This coin doesn't exist in real life...or would at least take some serious engineering to get exactly right...
Click to expand...

nor does an "exactly" fair coin with probabilities .5 and .5
 
Originally posted by: DrPizza
Originally posted by: b0mbrman
Originally posted by: alexjohnson16
Are you talking like a weighted coin?

I don't catch you.
Click to expand...
This coin doesn't exist in real life...or would at least take some serious engineering to get exactly right...
Click to expand...

nor does an "exactly" fair coin with probabilities .5 and .5
Click to expand...
Yup....I just call the ones that are close "coins"
 
Originally posted by: b0mbrman
Originally posted by: amdfanboy
.1 *.9 *.9 *.9 ..... ?
Click to expand...
That's the probability that he gets heads on the first flip and tails the next nine....

you want the sum of the probability of all ten permutations...so basically 10 times what you wrote

P(H-T-T-T-T-T-T-T-T-T-T) + P(T-H-T-T-T-T-T-T-T-T-T) + P(T-T-H-T-T-T-T-T-T-T-T) + ... + P(T-T-T-T-T-T-T-T-T-T-H)
=(0.1 x 0.9^9) + (0.9 x 0.1 x 0.9^8) + (0.9^2 x 0.1 x 0.9^7) + ... + (0.9^9 x 0.1)
= 10 (0.1 x 0.9^9)
=0.9^9
=pull out your calculator
Click to expand...

Isn't that what he wanted ?
 
Originally posted by: b0mbrman
Originally posted by: amdfanboy
.1 *.9 *.9 *.9 ..... ?
Click to expand...
That's the probability that he gets heads on the first flip and tails the next nine....

you want the sum of the probability of all ten permutations...so basically 10 times what you wrote

P(H-T-T-T-T-T-T-T-T-T-T) + P(T-H-T-T-T-T-T-T-T-T-T) + P(T-T-H-T-T-T-T-T-T-T-T) + ... + P(T-T-T-T-T-T-T-T-T-T-H)
=(0.1 x 0.9^9) + (0.9 x 0.1 x 0.9^8) + (0.9^2 x 0.1 x 0.9^7) + ... + (0.9^9 x 0.1)
= 10 (0.1 x 0.9^9)
=0.9^9
=pull out your calculator
Click to expand...

^ is correct.
 
Originally posted by: amdfanboy
Isn't that what he wanted ?
Click to expand...
It's one thing that would have satisfied the rule....but if at first he flipped a tail, then flipped a head, then flipped the rest tails...he still had "1 head AND 9 tails by flipping an UNFAIR coin...."

 
Originally posted by: amdfanboy
Originally posted by: b0mbrman
Originally posted by: amdfanboy
.1 *.9 *.9 *.9 ..... ?
Click to expand...
That's the probability that he gets heads on the first flip and tails the next nine....

you want the sum of the probability of all ten permutations...so basically 10 times what you wrote

P(H-T-T-T-T-T-T-T-T-T-T) + P(T-H-T-T-T-T-T-T-T-T-T) + P(T-T-H-T-T-T-T-T-T-T-T) + ... + P(T-T-T-T-T-T-T-T-T-T-H)
=(0.1 x 0.9^9) + (0.9 x 0.1 x 0.9^8) + (0.9^2 x 0.1 x 0.9^7) + ... + (0.9^9 x 0.1)
= 10 (0.1 x 0.9^9)
=0.9^9
=pull out your calculator
Click to expand...

Isn't that what he wanted ?
Click to expand...

sounds more like he wanted to know the probability, without any given order
 
Originally posted by: gopunk
Originally posted by: amdfanboy
Originally posted by: b0mbrman
Originally posted by: amdfanboy
.1 *.9 *.9 *.9 ..... ?
Click to expand...
That's the probability that he gets heads on the first flip and tails the next nine....

you want the sum of the probability of all ten permutations...so basically 10 times what you wrote

P(H-T-T-T-T-T-T-T-T-T-T) + P(T-H-T-T-T-T-T-T-T-T-T) + P(T-T-H-T-T-T-T-T-T-T-T) + ... + P(T-T-T-T-T-T-T-T-T-T-H)
=(0.1 x 0.9^9) + (0.9 x 0.1 x 0.9^8) + (0.9^2 x 0.1 x 0.9^7) + ... + (0.9^9 x 0.1)
= 10 (0.1 x 0.9^9)
=0.9^9
=pull out your calculator
Click to expand...

Isn't that what he wanted ?
Click to expand...

sounds more like he wanted to know the probability, without any given order
Click to expand...

Yup, order isnt important in this case. Thanks for the help.

 
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