You roll two fair dice. Let E be the event that an even total shows on the dice. Let F be the event that a three shows on at least one die. Find P(F) and P(F∣E).
To find the probability of event F, which is the event that a three shows on at least one die, we need to count the number of favorable outcomes and divide it by the total number of outcomes.
Step 1: Determine the favorable outcomes for event F.
We have two dice, and each die has six possible outcomes (numbers 1 to 6). To find the favorable outcomes, we need to consider the cases where a three shows on at least one die:
- The first die shows a 3, and the second die can be any number from 1 to 6. This gives us (1 option for the first die) * (6 options for the second die) = 6 favorable outcomes.
- The second die shows a 3, and the first die can be any number from 1 to 6. This also gives us 6 favorable outcomes.
Therefore, the total number of favorable outcomes for event F is 6 + 6 = 12.
Step 2: Determine the total number of outcomes.
When rolling two dice, and assuming they are fair, each die can have 6 possible outcomes. Since we have two dice, the total number of outcomes is 6 * 6 = 36.
Step 3: Calculate the probability of event F.
The probability of event F occurring is the number of favorable outcomes divided by the total number of outcomes:
P(F) = Favorable outcomes / Total outcomes
= 12 / 36
= 1/3
So the probability of event F, which is the event that a three shows on at least one die, is 1/3.
To find the conditional probability of event F given event E, denoted as P(F|E), we need to calculate the probability of F occurring given that E has occurred.
Step 1: Determine the favorable outcomes for event F and E occurring together.
Since E is the event of an even total, we need to consider the cases where a three shows on at least one die among the possibilities of even totals (2, 4, 6, 8, 10, 12).
To have an even total, the sum of two dice must be one of these even numbers.
The favorable outcomes for both events F and E are:
- The first die shows a 3, the second die shows an even number (2, 4, 6), or vice versa. This gives us 2 options.
So, the number of favorable outcomes for both F and E is 2.
Step 2: Determine the total number of outcomes given event E.
Since E is the event of an even total, we need to consider all the possible outcomes when rolling two dice and getting an even total. We have six even totals (2, 4, 6, 8, 10, 12), and the total number of outcomes for event E is 6.
Step 3: Calculate the conditional probability of event F given event E.
The conditional probability of F given E, denoted as P(F|E), is the number of favorable outcomes for both F and E divided by the total number of outcomes for E:
P(F|E) = Favorable outcomes for F and E occurring together / Total outcomes for E
= 2 / 6
= 1/3
So, the probability of event F (a three shows on at least one die) is 1/3, and the conditional probability of event F given event E (an even total shows on the dice) is also 1/3.
To find P(F), we need to determine the probability that a three shows on at least one die.
There are 36 possible outcomes when rolling two fair dice (6 possible outcomes for each die). Out of these 36 outcomes, there are 11 outcomes in which a three shows on at least one die:
(1, 3), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 3), (5, 3), (6, 3).
Therefore, P(F) = 11/36 ≈ 0.3056 or approximately 30.56%.
To find P(F|E), we need to determine the probability of occurring event F given that event E, an even total, has already occurred.
Out of the 36 possible outcomes, there are 18 outcomes in which an even total shows on the dice:
(1, 1), (1, 3), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (4, 2), (4, 4), (5, 5), (6, 2), (6, 4).
Out of these 18 outcomes, there are 5 outcomes in which a three shows on at least one die:
(1, 3), (3, 1), (3, 3), (4, 2), (6, 4).
Therefore, P(F|E) = 5/18 or approximately 0.2778, which is approximately 27.78%.
So, the probability of event F occurring is 0.3056 or 30.56%, and the probability of event F given that event E has occurred is 0.2778 or 27.78%.
p(f)=1/6*5/6 + 1/6*5/6 + 1/6*1/6
3x
y3
33
if a 3 shows on one die, then to be even sum, the other die has to be odd.
P(fgivenE)=1/6*3/6+3/6*1/6 only two ways.