Answer:
P(X= k) = (1-p)^k-1.p
Step-by-step explanation:
Given that the number of trials is
N < = k, the geometric distribution gives the probability that there are k-1 trials that result in failure(F) before the success(S) at the kth trials.
Given p = success,
1 - p = failure
Hence the distribution is described as: Pr ( FFFF.....FS)
Pr(X= k) = (1-p)(1-p)(1-p)....(1-p)p
Pr((X=k) = (1 - p)^ (k-1) .p
Since N<=k
Pr (X =k) = p(1-p)^k-1, k= 1,2,...k
0, elsewhere
If the probability is defined for Y, the number of failure before a success
Pr (Y= k) = p(1-p)^y......k= 0,1,2,3
0, elsewhere.
Given p= 0.2, k= 3,
P(X= 3) =( 0.2) × (1 - 0.2)²
P(X=3) = 0.128
Answer:
7³
Step-by-step explanation:
Using PEMDAS, we see that E (which stands for exponents) comes before M (which stands for multiplication) and A (which stands for addition) so the first operation you should do is 7³.
The one on the right would be greater
Problem 1
Answer: Independent
The reason why is because each bag is separate from one another, so one event doesn't affect the other. If we know the result of what we pulled out of one bag, it doesn't change the probability of the other event.
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Problem 2
Answer: Dependent
Assuming you do not put the first card back, then the probability of picking a King on the second draw will be different than if you picked a King on the first draw. With all 52 cards in the deck, the probability of getting a king is 4/52 = 1/13. It changes to 4/51 after we picked out an ace for the first card (and didn't put that first card back).
