JEE Advanced 2019 Paper 2 · Mathematics · Probability · hard

A fair die is rolled repeatedly until the first six appears. Let X be the number of rolls required. Then the conditional expectation E[Xmid X is…

Numerical answer

A fair die is rolled repeatedly until the first six appears. Let

X

be the number of rolls required. Then the conditional expectation

E\left[X\mid X\text{ is even}\right]

equals ______. (Give the answer correct to two decimal places.)

Show answer and solution

Answer: 6.55

Here

P(X=k)=\frac16\left(\frac56\right)^{k-1}

. Let

r=\frac56

. Then

P(X\text{ even})=\frac16\sum_{j\ge1}r^{2j-1}=\frac{r}{6(1-r^2)}=\frac{5/36}{11/36}=\frac5{11}

. Next

E\left[X\,\mathbf 1_{X\text{ even}}\right]=\frac16\sum_{j\ge1}2j\,r^{2j-1}=\frac{2}{6r}\cdot\frac{r^2}{(1-r^2)^2}=\frac{r}{3(1-r^2)^2}=\frac{5/18}{121/1296}=\frac{360}{121}

. Dividing,

E[X\mid X\text{ even}]=\frac{360/121}{5/11}=\frac{72}{11}\approx6.5454

, which is

6.55

. A direct numerical summation of the first thousand even terms agrees.

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