Probability Question

Hi people! I'm new to probability and so while going through conditional probability I found the following problem which I solved by arguing directly without using any results of Conditional Probability ............unfortunately the answer wasn't given. The problem goes as:

There are 4 friends A,B,C, D and a referee. A puts a '+' or a '-' sign on a piece of paper and passes it to B who perhaps changes the sign and passes the slip to C, who may change the sign and passes it to D. D perhaps changes the sign and finally passes the slip to the referee. Suppose 2/3 is the prob. of writing a '+' by A and each of B,C,D changes the sign by prob 1/3 independently. If the referee notices a '+' sign on the slip, then find the prob that A wrote a '+' on the slip.

I would really appreciate if someone gave me the solution. Also assuming the same conditions, what would then be the ans, instead of 4 friends if n friends did the same thing.

#Combinatorics #MathProblem #Math

Easy Math Editor

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Comments

This is my first impression of how to solve it. I'm not sure my numbers are all right, but I'm fairly sure this should be the approach:

The chance that a referee saw '+' is the chance that it was '+' originally and was changed 0 or 2 times, or the chance that it was '-' originally and was changed 1 or 3 times.

The first of these has chance $2/3$ (chance that A wrote '+') times the sum of $(2/3)^3$ and $3\cdot(1/3)^2\cdot 2/3$ (chance that it was changed 0 times or 2 times). Thus, the chance that A wrote '+' and the referee saw '+' is equal to $28/81$ .

The second has chance $1/3$ (chance that A wrote '-') times the sum of $(1/3)^3$ and $3\cdot(2/3)^2 \cdot 1/3$ (chance that it was changed 3 times or 1 time). This chance is $12 / 81$ .

Thus, the total chance that the referee saw '+' is $40/81$ . The chance that the referee saw '+' and A wrote '+' is $28/81$ , so the chance that A wrote '+' given the referee saw '+' is $(28 / 81) / (40 / 81) = 7 / 10$ .(Note the general identity $P(A \ \mathrm{and}\ B) = P(A) \cdot P(B | A)$ , in this case $A$ is that the referee saw '+' and $B$ is the chance that A wrote '+.')

To generalize this, note that the chance that the number of changes is odd is 1 minus the chance that the number of changes is even. The only difficulty is calculating one without knowing the other. Since the chance of changing is not $1/2$ , the sum cannot be collapsed due to symmetry. If there are $n$ middlemen (in this case, $n = 3$ ), then the chance of no changes is given by the first, third, etc. terms in the binomial expansion of $(2/3 + 1/3)^n$ . The eventual answer will be the quotient of $2/3$ times the even change chance and the sum of $2/3$ times the even change chance plus $1/3$ times the odd change chance.

Alexander Bourzutschky - 7 years, 7 months ago

Thanks Alexander, thanks a lot!........My solution was HIGHLY incorrect!!

Kulkul Chatterjee - 7 years, 7 months ago

Good question man............

Jit Ganguly - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$