Question 1089736: When two dice are rolled, the probability of getting a sum of 9 is 1/9 and the probability of not getting a sum of 9 is 8/9. If two dice are rolled n times, the probability of getting a sum of 9 exactly k times can be found by using the binomial theorem nCk(1/9)^k (8/9)^(n-k).
If a dice is rolled 8 times, find the probability of getting a sum of 9 at least two times.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Define the following events
A = event of getting a sum of 9 exactly zero times.
B = event of getting a sum of 9 exactly one time.
C = event that either A or B happens (pick one; cannot pick both)
C = event of rolling a sum of 9 exactly zero times OR exactly one time (pick one; cannot pick both)
------------------------------------------------------
For event A, n = 8 and k = 0 which means
n C k = (n!)/(k!*(n-k)!)
8 C 0 = (8!)/(0!*(8-0)!)
8 C 0 = (8!)/(0!*8!)
8 C 0 = (1)/(0!*1)
8 C 0 = (1)/(1)
8 C 0 = 1
Which is then plugged into the formula given
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (8 C 0)*(1/9)^(0)*(1-1/9)^(8-0)
P(X = 0) = (1)*(1/9)^(0)*(1-1/9)^(8-0)
P(X = 0) = 0.38974434312894
The probability of event A occurring is approximately 0.38974434312894
------------------------------------------------------
Repeat the same steps for event B. The value n = 8 stays the same. The value of k is now k = 1.
n C k = (n!)/(k!*(n-k)!)
8 C 1 = (8!)/(1!*(8-1)!)
8 C 1 = (8!)/(1!*7!)
8 C 1 = (8*7!)/(1!*7!)
8 C 1 = (8)/(1!)
8 C 1 = (8)/(1)
8 C 1 = 8/1
8 C 1 = 8
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (8 C 1)*(1/9)^(1)*(1-1/9)^(8-1)
P(X = 1) = (8)*(1/9)^(1)*(1-1/9)^(8-1)
P(X = 1) = 0.38974434312894
The probability of event B occurring is approximately 0.38974434312894
------------------------------------------------------
Add up the probabilities for event A and event B:
0.38974434312894+0.38974434312894 = 0.77948868625789
The probability of event A or event B happening is roughly 0.77948868625789
This is equivalent to saying: the probability of event C happening is roughly 0.77948868625789
Subtract that result from 1
1-0.77948868625789 = 0.2205113137421
which is the final answer. Round this value however you need to.
We subtract from 1 to find the complement event. The idea is that either event C happens or it doesn't. If event C doesn't happen, then it leads to the event of "getting a sum of 9 at least two times"
There's roughly a 22.05% chance of rolling a sum of nine at least twice given that you rolled a total of eight times.
|
|
|
