Wrong Solution for the unedited question.
Unedited question:
A fair die(with face numbered 1, 2, ....., 6) is independently rolled repeatedly.Let X denote the number is seen & let Y denote the number of rolls till 3 is seen. Evaluate E(Y|X =2)?
A) 41/6
B) 6
C) 11/2
D) 19/3
E) 17/3.
Solution:
I am getting 6 as an APPROXIMATED answer.
This is just an okay method, most probably there must be better & more intuitive solutions to this problem.
X & Y here are random variables. They will contain some values from their ranges of allowed values.
X : The value of X is seen.
Y: The value of Y denotes the number of rolls till a 3 is seen.
For example:
X = 2 implies 2 is seen,
Y = 4 implies 3 appeared for first time in the 4th die roll.
Ranges of Random Variables X & Y
X : { 1, 2, 3, 4, 5, 6}
Y: {$1, 2, 3, \ldots\infty$}
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E[Y | X = 2]: the expected (or average) number of rolls till a 3 is seen, given that 2 is seen.
In other words, average number of rolls till 3 appears for the first time in repeated die rolls, given that 2 has been already appeared at least once.
From the definition of Expectation:
$\displaystyle E[Y = y|X = 2] =\sum_{y=1}^{\infty}
{y * P(Y = y | X = 2)}$;
Where is the probability that 3 appeared for the first time in yth die roll, given that 2 already has been appeared at least once in the first (y - 1) die rolls.
P( Y = 1|X = 2) = 0, since there is only one die roll, 3 can't appear after 2.
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Calculation of P(Y = y|X = 2):
From the definition of conditional probability, for y die rolls:
$P(Y = y | X = 2) = \frac{(P(Y = y) \cap P(X = 2))}{P(X = 2)}$
i.e. $P(Y = y | X = 2) = \frac{\textrm{Probability of first appearance of 3 in the yth roll and AT LEAST ONE appearance of 2 in the first (y-1) die rolls}}{\textrm{Probability of at least one appearance of 2 in y die rolls}}$
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Calculating Denominator: (Probability of at least one appearance of 2 in y die rolls)
$P(Denominator) = \frac{\textrm{Outcomes in which 2 is appearing at least once}}{\textrm{All possible outcomes}}$
$P(Denominator) = \frac{(C_{1}^{y} 6^{(y – 1)} – (y – 1))}{6^y}$
$\textrm{How } (C_{1}^{y} *6^{(y – 1)} – (y – 1))$
Out of y places chose any one place for 2(yC1), in the remaining (y – 1) places any of the 6 outcomes can be placed minus (y -1) repetitions.
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Calculating Numerator: (Probability of first appearance of 3 in the yth roll and AT LEAST ONE appearance of 2 in the first (y-1) die rolls)
$P(Numerator) = \frac{\textrm{Outcomes in which 2 is appearing at least once in first (y-1) rolls and 3 is appearing in the last roll}}{\textrm{All possible outcomes}}$
$P(Numerator) = \frac{C_{1}^{(y - 1)}5^{(y -2)} – (y-2)}{6^y}$
$\textrm{How } C_{1}^{(y - 1)}5^{(y -2)}– (y-2)$
3 is fixed on the yth place, out of remaining (y – 1) places chose a place for 2 ((y-1)C1) and fill any of the 5 Outcomes(all possible outcomes EXCEPT 3) in the remaining (y-2) places minus (y -2) repetitions.
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Now,
$P(Y = y | X =2) = \frac{\textrm{Probability of first appearance of 3 in the y^th roll and AT LEAST ONE appearance of 2 in the first (y-1) die rolls}}{\textrm{Probability of at least one appearance of 2 in y die rolls}}$
$P(Y = y | X = 2) = \frac{P(Numerator)}{P(Denominator)}$
On putting the values & simplification,
$P(Y = y | X = 2) = \frac{((y-1)5^{(y-2)}-y+2)}{(y6^{(y-1)}-y+1)}$
By putting the value calculated above, Expectation E[Y | X = 2], can be given as follow:
$ \displaystyle E[Y = y|X = 2] = \sum_{y=2}^{\infty}
y*\frac{((y-1)5^{(y-2)}-y+2)}{(y*6^{(y-1)}-y+1)}$
I don’t know how to converge the Probability Mass Function(P(Y = y|X =2) so I calculated expectation by varying y from 2 to 120(Not Infinity)using fx991ES calculator & got 5.981334741.
I guess, Expectation to be 6.
Here is the plot for a portion of the Probability Mass Function(P(Y = y|X = 2) in which y is varying from 2 to 40, & weighted average of such a complete graph will give the expectation.