Suppose odd numbers 1, 3, 5 with probability c and on each of the even numbers with probability 2c.

(a) find c.
(b) suppose that the die is tossed. let x equal 1 if the result is an even number, and let it be 0 otherwise. also, let y equal 1 if the result is a number greater than three and let it be 0 otherwise. find the joint probability mass function of x and y. suppose now that 12 independent tosses of the die are made.
(c) find the probability that each of the six outcomes occurs exactly twice.

Step-by-step explanation:

a) Odd numbers if n then even numbers also n

Total prob = 1 gives 3c=1

c=1/3

b) X=1, if even and

    0, if odd

Y =1, if x >3 and

  =0 otherwise

xy can take values as 0 or 1

xy =1 if x even and y is greater than 3, or favorable outcomes are

4, 6.

xy= 1 if die shows 4 or 6

  =0 if otherwise

Prob (xy) = 1/3, for xy =1 and

               = 2/3 for xy =0

This is binomial since each trial is independent.

If 12 tosses are made then if we have binomial proba

with n =12 and p = 1/3 for success

what does x eaqual

step-by-step explanation:

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: