help with probability

Test Preparation
1

When Peter plays a game of darts the chances that he hits a bullseye is 0.65. Assume that each throw is independent.
What are the chances that three darts fired in succession will all hit bullseye?
What is the probability that none will hit?
What is the probability that at least one will hit?
What is the probability that 2 will hit?

2

All hit. There’s only one possibility - HHH (H=hit). Therefore, 0.650.650.65=0.275

None hit. One possibility - MMM (M=miss). And probability of missing is 1-0.65=0.35. Therefore, 0.350.350.35=0.043

At least one hit means that either 1, 2 or 3 hit. It’s easiest to do this as 1 - (probability that none hit)=1-0.043=0.957.
However, you can also do it by adding up the probabilities of 1, 2 and 3 hitting.
This would be (0.650.350.353)+(0.650.650.353)+(0.650.650.65)=0.957
Multiplying by three on the first two terms is because for 1 and 2 hits, there are three possibilities each: HMM, MHM, MMH and HHM, HMH, MHH.

Finally, the probability that two will hit, as mentioned above has three possibilites: HHM, HMH, MHH.
Therefore, it’s 0.650.650.35*3=0.444

3

thank you @unready

4

For the new SAT, you should also know the binomial probability formula.

  1. There are a fixed number of trials (n trials).
  2. For each trial, there are only two possible outcomes (A and not-A).
  3. Each trial is identical and independent.
  4. For each trial, the probability of the first outcome A is always the same (P(A)=p), and the probability of the second outcome not-A is always same (P(not-A))=1-p.

Then the probability that A occurs exactly k times out of n trials is:

nCk §^k(1-p)^(n-k)