Answer:
0.0788
Explanation:
In this case the final probability will be the sum of the probability of when there are 67 more reserves when there are 68.
That is to say:
P (f) = P (x = 67) + P (x = 68)
n = 68; p = 0.94
Now we calculate each one:
P (x) = nCx * (p ^ x) * [(1 - p) ^ (n-x)]
nCx = n! / (x! * (n - x)!)
Knowing the formula, we replace:
P (67) = 68C67 * (0.94 ^ 67) * [(1 - 0.94) ^ (68-67)]
nCx = 68! / (67! * (68 - 67)!) = 68! / 67! = 68
P (67) = 68 * (0.0158) * (0.06) = 0.064
Now for x = 68
P (68) = 68C68 * (0.94 ^ 68) * [(0.0641 - 0.94) ^ (68-68)]
nCx = 68! / (68! * (68 - 68)!) = 1
P (68) = 1 * (0.0148) * (1) = 0.0148
Then replacing in the main formula:
P (f) = P (x = 67) + P (x = 68)
P (f) = 0.064 + 0.0148
P (f) = 0.0788