A grocery store is running a sales promotion where a customer will receive one o

Question

A grocery store is running a sales promotion where a customer will receive one of the letters

A, E, L, S, U and V for each purchase. The letters are given away randomly by a cashier.

If a customer collects all six letters ("VALUES"), he/she will get a coupon for $10. What

is the expected number of purchases needed to get a coupon. (Hint: Let X be the number

of purchases needed to collect all six letters. Let Xi be the number of purchases to get the

ith missing letter, for i = 1, . . .. . . , 6, i.e., X = X1 +X2 + +X6. To nd the corresponding

probabilities, notice that the shopper always needs to get a letter he does not have yet. For

X1, any letter is `good', for X2, only 5 out of 6 letters will do, for X3, only 4 out of 6 will

give him what he needs to complete VALUES, etc.)

Explanation / Answer

A customer needs to procure all the six letters A,E,L,S,U and V at least once to get the coupon.

Now , the customer gets a coupon( whatever it be of A,E,L,S,U and V )at the first purchase and it is obviously different at So X1 = 1 with probability 1.

Next, he/she needs to get a different coupon.

Let X2 be the required no. of purchases made to get the a different letter.

In one purchase probability of getting a new letter is p1=5/6 .

So, X2 ~ Geometric(5/6) [ where X1 is the no. of trials And not denoting no. of failures before the success ].

So, E(X2) = 1/(5/6) = 6/5 .

Similarly when we get two different letters at least one each, X3 will be the no. of purchases to et the next distinct letter with probability 4/6 of getting in one purchase thereafter . So, X3 ~ Geometric(4/6).

And E(X3) = 1/(4/6) = 6/4 .

THis way , X4 ~ Geometric(3/6), X5 ~ Geometric(2/6) ,X6 ~ Geometric(1/6) .

and E(X4) = 1/(3/6)= 6/3 , E(X5) = 1/(2/6) = 6/2 , E(X6) = 1/(1/6) = 6.

So, E(X1 + X2 + X3 + ... + X6 ) = 6/6 + 6/5 + 6/4 + .....+ 6/1 = 6*(1/6 + 1/5 + 1/4 + ... + 1/1)

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