Suppose you are playing a game that requires rolls of two 4-sided dice, which have sides numbered 1 2, 3 and 4. (a) If both dice are fair, then what is the probability distribution over the set of possible outcomes for the sum of the two dice, when rolled? (Nit-picky note: a 4-sided die is a pyramid-looking thing; the result of a roll is the side facing down.) (b) Now suppose that one die is weighted such that a 3 is twice as likely to be rolled as any other single number on that die, and 1, 2 and 4 are all equally likely to be rolled. What is the probability distribution for outcomes of a single roll of just that one loaded die (c) What is the probability that the total of the fair and unfair dice is 6, when both are rolled? (d) Suppose you roll both dice but one falls on the floor and rolls under the couch so you can't see it, but the other die has rolled a 3. The unfair die is made of a strange material, which makes it twice as likely as the fair die to fall off the table. Without seeing the die that is under the couch, what is the probability that the dice total is 6? (e) Suppose these fair and unfair 4-sided dice, and two regular 6-sided dice, are all in a bag. If you pick a die at random from the bag and roll it, what it the probability of rolling a 3?

a) The probability distribution over the set of possible outcomes for the sum of the two dice.

x = single possible outcome

px = probability of the possible outcome

x px

2 (1/16) = 0.0625

3 (1/8) = 0.125

4 (3/16) = 0.1875

5 (1/4) = 0.25

6 (3/16) = 0.1875

7 (1/8) = 0.125

8 (1/16) = 0.0625

b) For just that one unfair die

x = single possible outcome

px = probability of the possible outcome

x px

1 (1/5) = 0.20

2 (1/5) = 0.20

3 (2/5) = 0.40

4 (1/5) = 0.20

c) The probability that the total of the fair and unfair dice is 6, when both are rolled = (1/5) = 0.20

d) Probability of getting a 6 in total given that the die on the table already gives a 3 = 0.35

e) Probability of rolling a 3 after picking a die at random from the bag and rolling it = (59/240) = 0.245833

Step-by-step explanation:

Each number on each four faced fair die has an equal probability of occurring = (1/4) = 0.25

For the sum of values of the dice

For 2; sample point is (1,1)

P(2) = (1/4) × (1/4) = (1/16)

For 3, sample points include (1,2) (2,1)

P(3) = [(1/4) × (1/4)] + [(1/4) × (1/4)] = (1/8)

For 4, sample points include (1,3) (2,2) (3,1)

P(4) = [(1/4) × (1/4)] + [(1/4) × (1/4)] + [(1/4) × (1/4)] = (3/16)

For 5, sample points include (1,4) (2,3) (3,2) (4,1)

P(5) = [(1/4) × (1/4)] + [(1/4) × (1/4)] + [(1/4) × (1/4)] + [(1/4) × (1/4)] = (1/4)

For 6, sample points include (2,4) (3,3) (4,2)

P(6) = [(1/4) × (1/4)] + [(1/4) × (1/4)] + [(1/4) × (1/4)] = (3/16)

For 7, sample points include (3,4) (4,3)

P(7) = [(1/4) × (1/4)] + [(1/4) × (1/4)] = (1/8)

For 8; sample point is (4,4)

P(8) = (1/4) × (1/4) = (1/16)

b) If the second die is an unfair die with 3 twice as likely to show up than others.

Let the probability of other numbers on the unfair die be x

The probability of getting a 3 on the unfair die = 2x

x + x + 2x + x = 1

5x = 1

x = (1/5)

2x = (2/5)

Probability of getting a 3 on the unfair die = (2/5)

Probability of getting any of the other number = (1/5)

c) For the sum of values of the fair and unfair dice.

Note that Probability of getting a 3 on the unfair die = (2/5)

And Probability of getting any of the other number = (1/5)

For 2; sample point is (1,1)

P(2) = (1/4) × (1/5) = (1/20)

For 3, sample points include (1,2) (2,1)

P(3) = [(1/4) × (1/5)] + [(1/4) × (1/5)] = (1/10)

For 4, sample points include (1,3) (2,2) (3,1)

P(4) = [(1/4) × (2/5)] + [(1/4) × (1/5)] + [(1/4) × (1/5)] = (1/5)

For 5, sample points include (1,4) (2,3) (3,2) (4,1)

P(5) = [(1/4) × (1/5)] + [(1/4) × (2/5)] + [(1/4) × (1/5)] + [(1/4) × (1/5)] = (1/4)

For 6, sample points include (2,4) (3,3) (4,2)

P(6) = [(1/4) × (1/5)] + [(1/4) × (2/5)] + [(1/4) × (1/5)] = (1/5)

For 7, sample points include (3,4) (4,3)

P(7) = [(1/4) × (1/5)] + [(1/4) × (2/5)] = (3/20)

For 8; sample point is (4,4)

P(8) = (1/4) × (1/5) = (1/20)

d) Suppose you roll both dice but one falls on the floor and rolls under the couch so you can't see it, but the other die has rolled a 3. The unfair die is made of a strange material, which makes it twice as likely as the fair die to fall off the table. Without seeing the die that is under the couch, what is the probability that the dice total is 6?

Let the Probability of the unloaded fair die falling off the table = x

Then, the Probability of the loaded unfair die falling off the table = 2x

x + 2x = 1

x = (1/3)

Probability of the unloaded fair die falling off the table = (1/3)

Probability of the loaded unfair die falling off the table = (2/3)

For a dice total of 6, given one die gives 3, the other die too has to give 3

Probability of getting a 6 in total is a sum of probabilities

1) Probability that the unfair die falls off to give the 3 given that the fair die stays on the table to give 3 = [(2/3) × (2/5)] = (4/15) = 0.26667

2) Probability that the fair die falls off to give the 3 given that the unfair die stays on the table to give 3 = [(1/3) × (1/4)] = (1/12) = 0.08333

Probability of getting a 6 in total given that the die on the table already gives a 3 = 0.26667 + 0.08333 = 0.35

e) Suppose these fair and unfair 4-sided dice, and two regular 6-sided dice, are all in a bag. If you pick a die at random from the bag and roll it, what it the probability of rolling a 3?

This is a sum of the Probabilities of getting a 3 on each die.

Probability of picking anyone of the four dice = (1/4)

The probability of rolling a 3 for each of the 4 dice are

Fair four-sided die; P(3) = (1/4)

Unfair four-sided die; P(3) = (2/5)

Six-sided fair die; P(3) = (1/6)

Six-sided fair die; P(3) = (1/6)

Probability of picking each die and getting a 3

Fair four-sided die; P(3) = (1/4) × (1/4) = (1/16)

Unfair four-sided die; P(3) = (1/4) × (2/5) = (1/10)

Six-sided fair die; P(3) = (1/4) × (1/6) = (1/24)

Six-sided fair die; P(3) = (1/4) × (1/6) = (1/24)

Total probability of picking a die and getting a 3 = (1/16) + (1/10) + (1/24) + (1/24) = (59/240) = 0.245833

his chances of getting a coin of higher value is 4/19

step-by-step explanation: