The probability mass function given a new random variable

Question

Say the random variable $Y = X^2 + 4$.

And $P(X = -2) = 1/10, P(X = -1) = 2/5, P(X = 0) = 1/4, P(X = 1) = 1/5, P(X = 2) = 1/20$

How would you find P(Y = 8)?

Additional question:

Say W is the number shown on a biased six-sided die. We know that P(W=w)=0.2 if w is an even number, and also that the probability the die shows a prime number is 0.5. What are the steps to finding the $P(X=1)$?

Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. — Shaun, Dec 23 '22 at 17:46
@TingoHugo Just set $Y=8$ and then solve the equation for $X$: $8=X^2+4\Rightarrow X^2=4\Rightarrow X=2 \cup X=-2$. So $P(Y=8)=P(X=-2)+P(X=2)$. — callculus42, Dec 23 '22 at 18:13
@callculus42 Can I trouble you with another question? Say W is the number shown on a biased six-sided die. We know that P(W=w)=0.2 if w is an even number, and also that the probability the die shows a prime number is 0.5. What are the steps to finding the P(X=1,2, etc)? — Tingo Hugo, Dec 23 '22 at 18:21
At the moment, with these information I cannot solve this question. It is asked for all values of $P(X=k)?$ — callculus42, Dec 23 '22 at 19:02

score 1 · Accepted Answer · answered Dec 23 '22 at 20:19

Prime numbers up to 6 are $2,3,5$. So $P(X=2)+P(X=3)+P(X=5)=0.5$. And we know that $P(X=2)=0.2$. Thus $P(X=3)+P(X=5)=0.3$.

So we know the following probabilities:

$P(X=2)=P(X=4)=P(X=6)=0.2$ and $P(X=3)+P(X=5)=0.3$

And we also know that the probability that any number is rolled is 1, which means that

$P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=1$.

Finally we can insert the values to calculate $P(X=1)$.

The probability mass function given a new random variable

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