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umka21 [38]
3 years ago
12

A four-sided die is loaded in a way that the probability of a face is proportional to the number of dots (there are 1, 2, 3, and

4 dots on the respective sides). (a) Construct a probability model that describes the likelihood of the outcomes occurring on one throw (that is determine the probability of 1, 2, 3, and 4 dots respectively). (b) Use a tree to determine the probability of getting a sum of 6 or more on two throws. Compare this to the probability if the die was fair.
Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0

Answer:

a) The probabilities are:

Pr(1) = 1/4

Pr(2) = 1/4

Pr(3) = 1/4

Pr(4) = 1/4

b) Pr(sum of 6 or more on two throw) = 6/16 = 3/8.

c) If the die was fair, it should have six (6) numbers => (1,2,3,4,5,6) and the sample space from two throw will be 6^{2} = 36.

Hence, Pr(sum of 6 or more on two throw) = 26/36 = 13/18.

Step-by-step explanation:

a) Since the probability of a face is proportional to the number of dots (there are 1, 2, 3, and 4 dots on the respective sides), then the occurrence of any outcome in a single throw are equally likely. Hence, there is equal probability of occurrence.

By probability, we say

Pr(x) = number of favorable event/ total number of possible events

Therefore, the total number = 4, and each can occur at equal probability. This is why the answer are:

Pr(1) = 1/4

Pr(2) = 1/4

Pr(3) = 1/4

Pr(4) = 1/4

b) The sample space when the die is thrown twice = 4^{2} = 16 possible outcome.

Among the outcomes, those that will have 6 or more when sum are just six (6). They are:

(2,4), (3,3), (3,4), (4,2), (4,3), and (4,4)

Thus, Pr(sum of 6 or more on two throw) = 6/16 = 3/8.

Just for understanding:

Other outcome from the sample space are:

(1,1), (1,2), (1,3), (1,4)

(2,1), (2,2), (2,3), ___

(3,1), (3,2), ___,  ___

(4,1), ___, ___,   ___

If you observe, the dash are the ones we have listed above!

c) When the die is fair, this is what happen.

Just for understanding:

Other outcome from the sample space when the die is fair are:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

The bold outcome are values when sum are 6 or more. And there are 26 of them.

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