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Two six sided fair dice are rolling. If sum of fallen numbers is less than $5$, than coin is flipping two times. Random variable $X$ represents sum of fallen numbers, $Y$ represents number of fallen tails. Find distribution of random variable $(X, Y)$. What is for example $P(X=2, Y=0) = ?$ I little confused, whenever I think of something, sum of all probabilities is less than $1$.

I made table, where $X = \{2,3,4,5,6,7,8,9,10,11,12\}$ $Y=\{0,1,2\}$ Than I filled it like I already told you: $P(X=2, Y=0) = \dfrac{1}{36}\cdot \dfrac{1}{4}$, $P(X=2, Y=1) = \dfrac{1}{36} \cdot \dfrac{2}{4}$, and so on. For $X \gt 5$, there are all zeros.

When I put probabilities for $X$, when $X\gt5$, sum of table is $2.44$ :( But isn't it logic that probabilities when $X \gt 5 = 0$, because you multiply with probability of getting Tails which is $0$ ? Also, when I assume that if $X\gt5$ coin is flipped $1$ time, I get sum $0.625$, which again is not right.

Lord_Farin
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user2982390
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3 Answers3

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The sum of all probabilities should be $1$.

If $X \geq 5$, then $Y = 0$ and so $X + Y = X$.

For example, if $X = 5$, then $Y = 0$, and we get $P(5,0) = (2/36)(1) = 1/18$.

Note also that $P(5, 1) = P(5, 2) = 0$.

For $X < 5$, you have:

$P(2, 0) = (1/36)(1/4), P(2, 1) = (1/36)(1/2), P(2, 2) = (1/36)(1/4)$.

Note the above line sums to $P(2,Y) = 1/36$.

$P(3, 0) = (2/36)(1/4), P(3, 1) = (2/36)(1/2), P(3, 2) = (2/36)(1/4)$.

Note the above line sums to $P(3, Y) = 2/36 = 1/18$.

$P(4, 0) = (3/36)(1/4), P(4, 1) = (3/36)(1/2), P(4, 2) = (3/36)(1/4)$.

Note the above line sums to $P(4, Y) = 3/36 = 1/12$.

If you let $Q(X)$ be the probability that you sum to $X$ after two rolls, then:

$P(2,0) + P(2,1) + P(2,2) = Q(2)$.

Similarly, $P(3,0) + P(3,1) + P(3,2) = Q(3)$ and $P(4,0) + P(4,1) + P(4,2) = Q(4)$.

For $X \geq 5$, we have $Q(X) = P(X,0) + P(X,1) + P(X,2) = P(X,0)$.

Thus, when you sum up all $P(X,Y)$ you get the same total as summing $Q(X)$, which is $1$.

I'm not quite sure how your $2.44$ appeared.

Benjamin Dickman
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$P\left\{ X=x\wedge Y=0\right\} =P\left\{ Y=0\mid X=x\right\} P\left\{ X=x\right\} $.

Here $P\left\{ Y=0\mid X=x\right\} =1$ if $x\geq5$ (no flipping coin) and $P\left\{ Y=0\mid X=x\right\} =\frac{1}{4}$ otherwise (coin flipped twice and no tails fall).

$P\left\{ X=x\wedge Y=1\right\} =P\left\{ Y=1\mid X=x\right\} P\left\{ X=x\right\} $.

Here $P\left\{ Y=1\mid X=x\right\} =0$ if $x\geq5$ (no flipping coin) and $P\left\{ Y=1\mid X=x\right\} =\frac{1}{2}$ otherwise (coin flipped twice and $1$ tail falls).

$P\left\{ X=x\wedge Y=2\right\} =P\left\{ Y=2\mid X=x\right\} P\left\{ X=x\right\} $.

Here $P\left\{ Y=2\mid X=x\right\} =0$ if $x\geq5$ (no flipping coin) and $P\left\{ Y=2\mid X=x\right\} =\frac{1}{4}$ otherwise (coin flipped twice and $2$ tails fall).

Next to that we have $P\left\{ X=x\right\}= \frac{1}{6}-\frac{\left|7-x\right|}{36}$.

drhab
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In the cases where you flip the coin $0$ times, you have a 100% probability of getting $0$ tails - thus, your table should have non-zero entries for each X,0 when $X\gt5$.

half-integer fan
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