Two fair dice are rolled. find the joint probability mass function of x and y when (a) x is the largest value obtained on any die and y is the sum of the values; (b) x is the value on the first die and y is the larger of the two values; (c) x is the smallest and y is the largest value obtained on the dice.

a)

P(X = x₀, Y = 2x₀) = 1/36

P(X = x₀, Y = k) = 1/18 for k between x₀+1 and 2x₀-1 inclusive

Every other event has probability 0. x₀ is any number between 1 and 6 inclusive.

b)

P(X = x₀, Y = x₀) = x₀/36

P(X = x₀, Y = k) = 1/36 for k between x₀+1 and 6 inclusive.

x₀ is between 1 and 6 inclusive. Every other event has probability 0.

c)

P(X = x₀, Y = x₀) = 1/36

P(X = x₀, Y = k) = 1/18 with k between x₀+1 and 6 inclusive

x₀ between 1 and 6 inclusive. Any other event has probability 0.

Step-by-step explanation:

Note that there are 36 possible results for the dice

a)

P(X = 1, Y = 2)

This is obtained only when both dices are 1, hence its probability is 1/36

P(X = 1, Y = k) = 0 (k > 1)

because if the largest value of the dice is 1, then both dices are 1

P(X = 2, Y = 3)

one dice is 2, the other one is 3, hence there are 2 possibilities and the probability is 2/36 = 1/18

P(X = 2, Y = 4)

This happens only if both dices are 2, hence the probability is 1/36.

P(X = 2, Y = k) = 0 (k > 2)

same argument of above. If the largest dice is 2, then the sum is either 3 or 4.

P(X = 3, Y = 4), P(X = 3, Y = 5)

in both given events we need one dice to be 3 and the other dice to be 1 for the first event and 2 for the second event. In both cases, there are only 2 favourable cases, hence the probability of the event is 2/36 = 1/18

P(X = 3, Y = 6)

This event happens only when both dices are 3, hence the probability is 1/36

This should show a pattern. As long as x₀ is between 1 and 6, if y₀ is between x+1 and 2x-1, then the probability P(X = x₀, Y = y₀) is 1/18 (either first dice is x₀, second dice is y₀-x₀ or first dice is x₀ and second dice is y₀ - x₀), also P(X = x₀, Y = 2x₀) = 1/36 (both dices are x₀). Every other event has probability 0.

b) We can separate them using conditional probability and the fact that both dices results are independent with each other.

P(X = x₀, Y = y₀) = P(X = x₀) * P(Y = y₀ | X = x₀)

P(X = x₀) = 1/6 for any value x₀ between 1 and 6.

If y₀ is x₀, this means that the first dice has the largest value, so the second dice is between 1 and x₀, and the probability of this event is x₀/6 (x₀ favourable cases over 6 possible ones).

If y₀ is not x₀, then it should be higher (otherwise the event would be impossible and it would have probability 0). As long as y₀ is between 2 and 6, the probability of this event is 1/6.

Thus

P(X = x₀, Y = x₀) = 1/6 * x₀/6 = x₀/36

P(X = x₀, Y = x₀ + k) = (1/6)² = 1/36 (k > 0)

Every other probability is 0

c)

P(X = x₀, Y = x₀) = 1/36 (because both dices are equal to x₀ in this event)

P(X = x₀, Y = x₀+k) = 2/36 = 1/18 (here k > 0. One possibility is the first dice is x₀ and the second one is x₀+k, and the remaining possibility is the first dice is x₀+k and the second dice is x₀)

Evert other event has probability 0.

(53.5 - 18.4) m.

step-by-step explanation:

after undergoing a displacement of 18.4m, the initial position was (53.5 - 18.4) m if an object has a final position of 53.5.

if i knew id you bro,, my bad