# Independent Probability Problem

If we repeat a same experiment in a laboratory independently 6 times. Evey time the probability that the experiment succeeds is 0.7.

a)What is the probability that it succeeds exactly 5 times?

b) What is the probability that it succeeds exactly twice?

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Opps been a long day. I edited above! Thank you for pointing out the error –  Q.matin Oct 2 '12 at 17:46
With your technique, would the probability that it succeeds exactly 0 times be 0.7/0? –  Did Oct 2 '12 at 17:47

Anytime that you have situation involving a sequence of $n$ independent trials that result in either success or failure, and you want to know the probability of exactly $k$ successes, you have what is called a binomial distribution. The probability of exactly $k$ successes out of $n$ trials with probability $p$ of success is given by

$$\begin{pmatrix} n\\k \end{pmatrix}p^k\left(1-p\right)^{n-k}.$$

Where $(~^n_k)$, read as "n choose k," is the number of ways that you can choose $k$ elements from a set of $n$ elements and equals $$\frac{n!}{k!(n-k)!}.$$ In your case, $n=6$, $p=0.7$, and for part (a) $k=5$ and for part (b) $k=2$.

The logic for this construction is as follows. I will describe the process for part (a), and the logic will be the same for part (b). We can calculate the probability $P(SSSSSF)=(0.7)^5 0.3$ using the multiplication rule for independent events, which is the probability of getting first five successes and then one failure, but this is not the only way get five successes. We could also get the series $(SSSFSS)$, so we need to add up the probability for all of these different possibilities that have the same probability.

The number of such events called $(~^6_5)$ which is the number of ways to choose five sucesses out of six trials. If you need help with knowing why this is true, you can ask in a comment and I will add it to my answer, or better yet search SE for an answer. We can therefore multiply $(0.7)^50.3$ and $(~^6_5)$ to get the answer.

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Thank you very much, Carl !! I understand the permutation you described above! And also, I go to Georgia Tech too! –  Q.matin Oct 2 '12 at 18:13
It's great to meet a fellow GT student on SE. I'm glad I could be of help! –  Carl Morris Oct 2 '12 at 18:15

We want the probability of exactly $5$ successes, and therefore $1$ failure, in $6$ trials.

The $5$ successes and $1$ failure could happen in several orders. They are $SSSSSF$, $SSSSFS$, $SSSFSS$, $SSFSSS$, $SFSSSS$, and $FSSSSS$,

Calculate the probability of each and add them up. For example, the probability of the pattern $SSSSSF$ is $(0.7)(0.7)(0.7)(0.7)(0.7)(0.3)$.

For (b), you will probably want to count the number of patterns more efficiently than by listing.

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Oh okay, so its just a permutation type problem? I calculated and got that each succesive trial equals 0.050421 and then I multiplied it by 5 so that equals to .2521? Is that correct? –  Q.matin Oct 2 '12 at 17:57
@Q.matin No. Multiplying by 5 is your mistake. Recount how many different ways 5 sucesses can happen in 6 trials. –  rschwieb Oct 2 '12 at 18:10

André has answered your question, but for the record: with independent events, an and translates to multiplication, an or translates to addition.

ex.: in a coin-toss scenario, the probability of getting 2 heads in 2 tosses (heads and heads) is $1/2\times 1/2=1/4$. The probability of getting either heads or tails (in one toss) is $1/2+1/2=1$, which makes perfect sense.

By applying this to your exercise, you should be able to figure out the answers. (don't forget to include the failure probabilities as well as the success probabilities)

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Thank you very much for the pointer! Just to verify I did my question correct,I calculated and got that each succesive trial equals 0.050421 and then I multiplied it by 5 so that equals to .2521. Is that correct? –  Q.matin Oct 2 '12 at 18:04
Multiply by 6, since 5 successes and 1 failure can occur in 6 different orders. –  user39572 Oct 2 '12 at 18:08
@Q.matin: If you count the number of ways in the list that I made, $SSSSSF$ and so on, you will see there are $6$. The lone failure can happen in any one of the trials. –  André Nicolas Oct 2 '12 at 19:11