I have this question and I got part a but I am confused where to start on question b

The probability distribution of a discrete random variable X is given by:

P(X = r) = k(6r2 −r3) for r = 1, 2, 3, 4, 5

P(X = r) = 0 otherwise

(a) Show that k = 1/105

Two values of X are chosen at random.

(b) Find the probability that the product of the two values is even.

The probability distribution of a discrete random variable X is given by:

P(X = r) = k(6r2 −r3) for r = 1, 2, 3, 4, 5

P(X = r) = 0 otherwise

(a) Show that k = 1/105

Two values of X are chosen at random.

(b) Find the probability that the product of the two values is even.

Original post by Magicmanzb

I have this question and I got part a but I am confused where to start on question b

The probability distribution of a discrete random variable X is given by:

P(X = r) = k(6r2 −r3) for r = 1, 2, 3, 4, 5

P(X = r) = 0 otherwise

(a) Show that k = 1/105

Two values of X are chosen at random.

(b) Find the probability that the product of the two values is even.

The probability distribution of a discrete random variable X is given by:

P(X = r) = k(6r2 −r3) for r = 1, 2, 3, 4, 5

P(X = r) = 0 otherwise

(a) Show that k = 1/105

Two values of X are chosen at random.

(b) Find the probability that the product of the two values is even.

You could list (tree/grid) the outcomes/probability of the product of two values and sum the ones that are even. Or do 1-odd probability.

(edited 3 years ago)

Original post by mqb2766

You could list (tree/grid) the outcomes/probability of the product of two values and sum the ones that are even. Or do 1-odd probability.

Thank you but how do I work out the values of X?

Original post by Magicmanzb

Thank you but how do I work out the values of X?

If you draw a tree, for instance, you have to consider the different values it can take (as with most probability problems).

If you consider only the pair's that generate even or odd, it cuts down the options a bit.

(edited 3 years ago)

Original post by mqb2766

If you draw a tree, for instance, you have to consider the different values it can take (as with most probability problems).

If you consider only the pair's that generate even or odd, it cuts down the options a bit.

If you consider only the pair's that generate even or odd, it cuts down the options a bit.

So do I use the numbers 1-5 on the tree?

Original post by Magicmanzb

So do I use the numbers 1-5 on the tree?

Those are the outcomes with a nonzero probability for one score.

A bit of thinking about the question may give an efficient way to get the final probability using even/odd, but failing that, just list the individual probabilities.

(edited 3 years ago)

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can someone please explain what principle domain is and why the answer is a not c?Maths

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