Question

A bag contains 3 coins, 1 is baised with both side heads and other two are normal.
Assume 'A' to be baised, ['B' and 'C'] non baised.
A coin is chosen at random from the bag and tossed 4 times. If heads turn up each time, what is the prob that it is a 2 headed coin.

Answer 1

put \(Y\) as the event the coin is two headed, \(X\) as the event you get 4 heads, and you need to compute
\[P(Y|X)=\frac{P(Y\cap X)}{P(X)}\]
each coin is chosen with probability \(\frac{1}{3}\)

Answer 2

I know i will have to use Baye's theorem in this...

prob that A, B or C turns up is P(A)=P(B)=P(C)=1/3
Prob that heads turns from A, B or C will be
P( H|A) =1

Answer 3

the real work is computing \(P(X)\)

Answer 4

I want to know why Prob of P( H|B) =\[(\frac{ 1 }{ 2 }) ^{4}\]

Answer 5

if it is an unbiased coin, then the probability you get 4 heads is \(\frac{1}{16}\) so the probability you get 4 heads is
\[\frac{1}{3}\times \frac{1}{16}+\frac{1}{3}\times \frac{1}{16}+\frac{1}{3}\times 1\]

Answer 6

the power 4 is there coz of coin being tossed 4 times??

Answer 7

i.e.
coin B is chosen, AND get 4 heads OR coin C is chosen AND get 4 heads OR coin A is chosen AND get 4 heads

Answer 8

yes, the probability you get 4 heads on an unbiased coin is \(\left(\frac{1}{2}\right)^4=\frac{1}{16}\)

Answer 9

numerator of your fraction is \(\frac{1}{3}\) and denominator is \(\frac{1}{3}\times \frac{1}{16}+\frac{1}{3}\times \frac{1}{16}+\frac{1}{3}\times 1\) whatever that is

Answer 10

OK can u suggest me a good book where I can find solution or solved examples for probability??

Answer 11

depends on what level you want
Feller used to be standard text
there is a good book by Galombos
and one on discrete probability that i like by Gordon
but I would look for free on line resources

Answer 12

I am 2nd year college student.

Answer 13

I am reffering Meyer

Answer 14

look on line

Answer 15

online as in Yahoo answers??

Answer 16

i like google
there are no doubt free on line texts

Answer 17

thank you