Question

Suppose you have 8 dice in a bag. You draw a single die and roll it. Three dice are standard fair dice, numbered 1-6 Two dice are have one less pip in the middle of the 5 side , such that there is no 5 on these two dice, but there are two 4's (that is, these dice are numbered {1, 2, 3, 4, 4, 6}) Three dice are ten-sided, and loaded such that it is twice as likely that an even number is rolled than an odd What is the probability that you roll a 4

Answer 1

Answer:

The probability of drawing a 4 is $\frac{47}{240}$

Step-by-step explanation:

We will assume that each of the dices is equally likely to be drawn. So, let us consider this three events: A is the event that we draw a traditional die, B is the event that we draw a die that has two 4's on it and C is that we draw a ten sided die.

Since they are all mutually exclusive events, we have that P(A) + P(B) + P(C)=1. We will find this probability by simply counting the number of ways in which we get the specific event and divided by the total number of outcomes.

Note that since we have 8 dice, and 3 dice are fair standard dice, P(A) = 3/8.

On the same fashion, we get that P(B) = 2/8, P(C) = 3/8.

No, we will calculate the probability of getting a 4 to each type of die. Let D be the event that we get a 4. So, we will calculate the conditionals probabilities P(D|A), P(D|B), P(D|C). Recall that P(D|A) is the probability of getting a 4, given that we draw a standard fair die.

So, suppose we get a standard die. Since it is fair and standard, we have that  the probability of getting a four is 1/6. Thus P(D|A) = 1/6.

If we get a die that has two 4's, we have double the chance of getting a 4, so the probability is 2/6. Then P(D|B) = 2/6.

Now, consider the case we get a 10-sided die. In here, we will assume that all odd numbers are equally likely between them and that even numbers are equally likely between them. When we throw the die, we get a number between 1 and 10. Since all possible outcomes of throwing the die once are mutually exclusive, we must have that P(1)+P(2)+...+P(10)= 1. Let c be the probability assigned to an odd number and 2c be the probability assigned to an even number. Then we have that

$c+ 2c+c+2c+c+2c+c+2c+c+2c = 5c+5*(2c) = 15c = 1$

which implies that c= 1/15. Then, in this case the probability of getting a 4 is 2c, i.e 2/15. Then P(D|C) = 2/15.

We will use the following theorem (total probability theorem). Given a partition of the sample space $A_1, \dots , A_n$ (partition means mutually exclusive events) and an event B, then

$P(B) = \sum_{i=1}^{n} P(A_i)P(B|A_i)$

In our case, we are asked to calculate P(D). Then, by this theorem

$P(D) = P(A)\cdot P(D|A)+P(B) P(D|B) + P(C) P(D|C) = \frac{3}{8}\cdot \frac{1}{6}+\frac{2}{8}\cdot \frac{2}{6}+\frac{3}{8}\cdot \frac{2}{15}= \frac{47}{240}$