- Conditional probability rolling two dice
- probability of rolling two dice: independent or not?
- When 3 dice are rolled simultaneously what is the probability that exactly two of the dice will come up as the same number?
Conditional probability rolling two dice
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? I already know the answer, but am having some trouble understanding it. If E is the event that at least one dice lands on 6 and F is the event that the dice land on different numbers, I need to calculate P(EF)/P(F) According to the answer P(EF) = 2*(1/6)*5/6. I don't understand how this was calculated. Is there a specific formula for calculating P(EF)? I know P(EF) = P(E)+P(F) if the two events are mutually exclusive. P(F) = 30/36 which makes sense because there are 36 possible outcomes, of which 30 are favorable. Mainly just confused about calculating P(EF). (SIde note: The events are neither mutually exclusive nor independent.) $P(E\cap F)$ is the probability that one die is a six and the other die is not. That's $\frac 1 6\frac 5 6+\frac 5 6\frac 1 6$ by adding the probability that the first die is a six and the other not, to the probability that the first die is not a six and the other is. (NB: Those events are mutually exclusive partitions of $E\cap F$.) $$\mathsf P(E\cap F) = 2 \cdot \frac 1 6 \cdot \frac 5 6\\ = \frac$$ It might be useful to use a symmetry argument to count the size of EF. EF has 30 pairs of numbers. That's a total of 60 numbers. Since each of the six choices shows up exactly the same number of times (there's the symmetry argument), 1/6*60 = 10 of those 60 numbers is the number '6'. And each of those 10 is in a di...
probability of rolling two dice: independent or not?
In researching a probability question, I found my answer in this old Math stackexchange question here: However, the answers posted assume that the dice rolls are independent events so that the probabilities can be multiplied. I was just curious if it makes a difference in the calculation if: the dice are rolled at the same time, or if they are rolled one after another? Thanks. There's no difference between the two procedures (throwing simultaneously or one-by-one) as far as independence is concerned. To see this you might go back to the formal definition of independence. Definition. (Statistical) independence of two events. Two events $A$ and $B$ are independent if their joint probability equals the product of their respective probabilities, i.e. $$P(A \cap B)=P(A)P(B)$$ As we see, it's NOT like independence (statistical independence) implies the product rule, rather the concept of statistical independence is defined by the product rule. It is easy to check from the joint probability distribution that throwing of two dices are statistically independent. Assume that the die is fair $($i.e. each of the sides come up with equal probability of $\frac = P(X=i)P(X=j),$$ for $i=1(1)6, ~j=1(1)6$. We can use this to compute $P(X \in A, ~Y \in B)$ which comes out to be $P(X \in A)P(Y \in B)$. Now check! Do you think that the joint distribution of $(X, Y)$ changes because the dice are thrown together or one-by-one? In fact, after the experiment is done, and the outcome is attached to...
When 3 dice are rolled simultaneously what is the probability that exactly two of the dice will come up as the same number?
There are a total of 6 x 6 x 6 = 216 outcomes. Let’s say the two numbers that come up are 1 and 2, and then we could have any of the following: (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 2, 2), (2, 1, 2) and (2, 2, 1) However, we have 6 C 2 = ( 6 × 5 ) / 2 = 1 5 ways to choose two distinct numbers from six. For each of these 15 pairs of numbers, we have 6 ways (see the example for 1 and 2 above). Therefore, we have a total of 15 x 6 = 90 ways that exactly two of the dice will come up as the same number. So the probability is 90/216 = 15/36 = 5/12.
