Question

There are 10 numbers in a hat, and you choose seven of them without replacement. Let X be the minimum of the seven numbers drawn, and Y be the maximum of the seven numbers drawn. a. Find the probability distribution for X.

b. Find the probability distribution for Y .

c. Find the expectation of Y X.

Answer 1

Let the set of 10 numbers in ascending order be (n₁, n₂,....., n₁₀) where n₁ is the lowest number and n₁₀ is the highest number.

Total no. of ways in selecting 7 Nos. out of 10 = = 120

(a) Let us consider the following mutually exclusive cases in selecting the 7 nos.,

including the lowest no. x₁ = = 84

including the 2nd lowest no. n₂ but not including n₁ = = 28

including the 3rd lowest no. n₃ but not including n₁ and n₂ = = 7

not including any of the n₁ , n₂ andn₃ (i.e. n₄ is the minimum no. in the selection) = = 1

Since, random variable X is the minimum of the 7 numbers drawn, it has following values (n₁, n₂, n₃, n₄) with probabilities ( , , , )

(b) Applying similar reasoning to the above,

  

Let us consider the following mutually exclusive cases in selecting the 7 nos.,

including the highest no. n₁₀ = = 84

including the 2nd highest no. n₉ but not including n₁₀ = = 28

including the 3rd highest no. n₈ but not including n₉ and n₁₀ = = 7

not including any of the n₈ , n₉ andn₁₀ (i.e. n₇ is the minimum no. in the selection) = = 1

Since, random variable Y is the minimum of the 7 numbers drawn, it has following values (n₇, n₈, n₉, n₁₀) with probabilities ( , , , )

(c) The expectation of YX will be

E(YX) = E(Y).E(X) since Y and X are independent.

and E(Y) = n_{7 .} (1/120) + n_{8 .} (7/120)+ n₉ . (28/120) + n_10. (84/120)

E(X) = n_1. (84/120) + n₂ . (28/120) + n_{3 .} (7/120) + n_{4 .} (1/120)