#probability, flipping coin

57 messages · Page 1 of 1 (latest)

raven delta
#

Hi, You have 2 coins, one is fair and the other one is with both sides heads. You flip one of the coins 3 times , if you know that you have gotten 3 heads what is the probability you chosen the fair coin?

wraith terrace
raven delta
#

yes, but im unsure how to apply it

wraith terrace
#

alright

#

so basically

#

you want to find the probility that each coin, if chosen, would give three heads in a row

#

if you flip the biased coin, it's ofc 100% chance

#

what about if you pick the fair coin? then what's the chance of three heads in a row?

raven delta
#

1/8 ?

wraith terrace
#

yeah nice

#

so that means the biased coin is 8 times more likely to have three heads in a row than the fair coin

raven delta
#

yes

wraith terrace
#

so the coin is 8 times more likely to be biased than fair

#

so if the probability of fair coin is x, and biased is 8x, and x+8x=1, what is x?

raven delta
#

1/9?

wraith terrace
#

heck yeah

raven delta
#

haha thats all?😅 , you dont have to use the bayers formula like P(H|T) or something?

wraith terrace
#

I mean you can but this is a little simpler imo

#

basically events X and Y happening both have 50% chance

#

...actually I was going to say something but it'd be too convoluted

raven delta
#

okay, idk i have a hard time knowing which one event should be the A or B in the bayers theorem. i hate statistics/probabilty😅

wraith terrace
#

yeah I like not using formulas whenever possible

raven delta
#

we are supposed to use it so thats why im trying to learn how to

wraith terrace
#

oh ok

#

well I can show you how to use bayes if you want

#

for this problem

raven delta
#

i would very much appreciate it

wraith terrace
#

sure

#

so we want P(fair coin | three heads in a row)

raven delta
#

yes

wraith terrace
#

which means using bayes that's equal to P(three heads | fair coin)* P(fair coin) / P(three heads in a row)

#

P(fair coin) is 1/2 since you just pick a coin randomly

#

P(three heads | fair coin) is 1/8 as we discussed

#

P(three heads in a row) = P (three heads & fair coin) + P(three heads & biased coin) = 1/2 + 1/16 = 9/16

#

so 1/8 * 1/2 / 9/16 = 1/9

raven delta
#

im just gonna write it down so i see it, but i seem to understand it

wraith terrace
#

awesome

raven delta
#

wow that felt more easier than i thought, thank you for ur help.

Just out of curiosity, if we would want to know what is P(HHH). in other words, the probability of getting three heads. The chances are different depending if u take a fair or biased coin. its 1 if the biased coin is taken, and 1/2 if fair coins is taken.

But could you calculate it first for the fair coin and then the biased one, and just add the probabilities together?

wraith terrace
#

probability of 3 heads is 1/8 with fair coin not 1/2

#

you do just add them but you have to multiply the value for each coin by 0.5 bc there's a 50% chance of either coin

raven delta
wraith terrace
#

just a heads up I gtg after this upcoming question

raven delta
wraith terrace
#

you want to find just P(HHH)?

#

then it's 1/2 * (HHH | fair) + 1/2 * (HHH | biased)

#

= 1/2 * 1/8 + 1/2 * 1

raven delta
#

ohh

wraith terrace
#

= 1/16 + 1/2 = 9/16

raven delta
#

okay

#

thank u

wraith terrace
#

no prob

#

I gtg but have a good one

raven delta
#

you too1

#

!*

wraith terrace
#

thanks!

rapid bramble
# raven delta Hi, You have 2 coins, one is fair and the other one is with both sides heads. Yo...

F: denote the event you picked the fair coin
B: denote the event you picked the biased coin
D: Data collected, i.e., observed two heads in two tosses

We want to calculate: P(F|D) and P(B|D)

Assume that the trials are independent. This is equivalent to saying:
P(D|F)=1/4 (product of 0.5 and 0.5)
P(D|B)=9/16

Let us also assume that P(F)=P(B)=0.5, implying that "you randomly pick coin" with equal chances of picking either one.

Apply Bayes theorem, which basically starts from:

P(F|D)P(D)=P(D|F)P(F)

Bring over the second term on the left hand side to the right hand side and expand by considering total probability as:

P(D)=P(D|F)P(F)+P(D|B)P(B) = 1/41/2+9/161/2

P(F|D) = [P(D|F)P(F)]/ [(P(D|F)P(F)+P(D|B)P(B))]

Please substitute and solve :-)
I got 4/13

wraith terrace