You roll a red number cube and a blue number cube. Let A be the event "at least one number cube shows a 6"
a) Find P(A) by finding the sum of P(6 on red cube, no 6 on blue cube), P(no 6 on red cube, 6 on blue cube), and P(6 on both cubes)
b) Describe the event "not A" then find P(not A) and use it to find P(A)
c) Compare the methods in part a and b.
Assuming both are fair cubes.
Any fair cube will give a "6" with theoretical probability of 1/6.
(a)
For case
Red cube: P(6)=1/6, P(~6)=5/6
Blue cube: P(6)=1/6, P(~6)=5/6
Using the multiplication rule for the two step experiment (red, then blue),
P(6R,~6B)=1/6*5/6
P(~6R,6B)=5/6*1/6
P(6R,6B)=1/6*1/6
Add the sum of probabilities of the three cases above.
(b) ~A=both cubes don't show a 6.
P(~A)=5/6*5/6
P(A)=1-P(~A)= ?
(c)
Do your comparison.
In probability, it is advantageous to work with fractions whenever possible because it will give unambiguous results.
I have no idea what the answer could possibly be
a) To find P(A), we need to find the probability of at least one of the cubes showing a 6.
Let's calculate each individual probability:
P(6 on red cube, no 6 on blue cube):
The probability of rolling a 6 on the red cube is 1/6, and the probability of not rolling a 6 on the blue cube is 5/6. Therefore, the probability of this event occurring is (1/6) * (5/6) = 5/36.
P(no 6 on red cube, 6 on blue cube):
The probability of not rolling a 6 on the red cube is 5/6, and the probability of rolling a 6 on the blue cube is 1/6. Therefore, the probability of this event occurring is (5/6) * (1/6) = 5/36.
P(6 on both cubes):
The probability of rolling a 6 on both the red and blue cubes is (1/6) * (1/6) = 1/36.
Now, to find P(A), we sum up these probabilities:
P(A) = P(6 on red cube, no 6 on blue cube) + P(no 6 on red cube, 6 on blue cube) + P(6 on both cubes)
= 5/36 + 5/36 + 1/36
= 11/36
b) The event "not A" refers to the scenario where neither cube shows a 6. So, P(not A) can be calculated as the probability of neither cube showing a 6.
Since the red cube has a 5/6 probability of not showing a 6, and the blue cube has a 5/6 probability of not showing a 6, we can calculate P(not A) as:
P(not A) = P(no 6 on red cube) * P(no 6 on blue cube)
= (5/6) * (5/6)
= 25/36
To find P(A) using P(not A), we can subtract P(not A) from 1 (1 - P(not A)):
P(A) = 1 - P(not A)
= 1 - (25/36)
= 11/36
c) Both methods in part a) and b) result in the same value for P(A), which is 11/36. The difference lies in how we approach calculating the probability. In part a), we directly calculate the probabilities of each event and sum them up to find P(A). In part b), we find P(not A) and use it to find P(A) by subtracting it from 1. Both methods are valid and should yield the same result.
a) To find P(A), we need to calculate the probabilities of the different outcomes that satisfy the condition "at least one number cube shows a 6".
Let's break it down into three cases:
Case 1: 6 on the red cube, no 6 on the blue cube. The probability of rolling a 6 on the red cube is 1/6, and the probability of not rolling a 6 on the blue cube is 5/6. Therefore, the probability of this case is (1/6) * (5/6) = 5/36.
Case 2: No 6 on the red cube, 6 on the blue cube. The probability of not rolling a 6 on the red cube is 5/6, and the probability of rolling a 6 on the blue cube is 1/6. Therefore, the probability of this case is (5/6) * (1/6) = 5/36.
Case 3: 6 on both the red and blue cubes. The probability of rolling a 6 on both cubes is 1/6 * 1/6 = 1/36.
Now, add up the probabilities of these three cases to find P(A):
P(A) = 5/36 + 5/36 + 1/36 = 11/36.
b) The event "not A" refers to the situation where both number cubes do not show a 6. In other words, "not A" means neither the red cube nor the blue cube shows a 6.
To find P(not A), we need to calculate the probability of not rolling a 6 on both the red and blue cubes.
The probability of not rolling a 6 on the red cube is 5/6, and the probability of not rolling a 6 on the blue cube is also 5/6.
Therefore, P(not A) = (5/6) * (5/6) = 25/36.
To find P(A) using P(not A), we can use the complement rule:
P(A) = 1 - P(not A) = 1 - 25/36 = 11/36 (same result as in part a).
c) Both methods yield the same result for P(A), which is 11/36. However, the approaches are different.
In part a, we directly calculated the probabilities of the different cases that satisfy the condition "at least one number cube shows a 6" and then added them up.
In part b, we calculated the probability of the complementary event, "not A," and used it to find P(A) using the complement rule.
Both methods are valid, but depending on the specific problem, one method may be easier or more straightforward than the other. In this case, since it is easier to calculate the probability of not rolling a 6 on both cubes, using the complement rule in part b is more convenient.