4.3 A basket contains 4 puppies: one of the puppies has 1 spot, one of the puppi

Question

4.3 A basket contains 4 puppies: one of the puppies has 1 spot, one of the puppies has 2 spots, and the remaining two puppies have 4 spots. Suppose two puppies are selected at random without replacement. Let the random variable X equal the total number of spots on the selected puppies. (a) Find the probability distribution of X (b) Find the probability that the puppies have a total of 5 spots, i.e. find P(X -5) (c) Find the probability that the puppies have a total of 6 or more spots, i.e. find P(X 26) (d) Find the probability that the puppies have 5 or fewer spots or 8 spots, i.e. find (e) Given that the puppies have 6 or more spots, determine the probability that both puppies have 4 spots each (i.e. 8 spots total), i.e. find P(X -8X 26) (f) On average, how many spots do we expect on the two selected puppies? (g) Compute 2-Var(X).

Explanation / Answer

In total there are 4 puppies, one puupy has 1 spot, one puppy has 2 spots and two puppies have 4 spots.

Two puppies are selected at random from the basket without replacement and The random variable X denotes the total number of spots on the selected puppies.

So, the possible value the random variable X can take are 1+2, 1+4, 2+4, 4+4 that is 3, 5, 6 and 8.

(a) The probability distribution of the random variable X can be given by:

(b) From the above table the probability that the selected puppies have a total of 5 spots is P(X=5) = 1/3

calculated as 1C1*2C1 / 4C2 =1/3

(c) The probability that the puppies will have 6 or more spots is P(X6) = P(X=6) +P(X=8) =1/3 + 1/6 =3/6 = 1/2

(d) The required probability is P(X5 or X=8) = P(X=3) + P(X=8) = 1/6 + 1/6 = 1/3

(e) The probability required to be calculated here is P(X=8|X6) = P(X=8 X6)/ P(X6) = P(X=8)/P(X6)

Using the result in part (c), P(X=8|X6) =(1/6)/(1/2) = 1/3

(f) The average number of spots in the selected puppies is given by the expectation:

i.e. E(X) = 3x1/6 + 5x1/3 + 6x1/3 + 8x1/6 = 5.5

Thus, on average 5.5(~6) spots are expected on the two puppies.

(g) The variance of the number of spot X is given by:

V(X) = E(X2)-[E(X)]2

Now, E(X2) = 32x1/6 + 52x1/3 + 62x1/3 + 82x1/6 = 32.5

Therefore, V(X) = 32.5 -5.52 = 2.25

X 3 5 6 8 P(X=x) 1/6 1/3 1/3 1/6

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