There are two questions that I am having problems with.

The first one is this one:

             Event A: an even number turns up

             Event B: a 5 or 6 turns up

A die is rolled once: The probability of getting a 1 is 1/21, a 2 is 2/21, a 3 is 3/21, a 4 is 4/21,a 5 is 5/21, and a 6 is 6/21.

They ask you to find P(A), P(B), P(AUB), P(AnB), P(AlB), P(BlA), and if the events are independent. I have an idea of how to do it and my results were:

P(A): 4/7 ---- P(b): 11/21 ----P(AnB): 6/21----- P(AUB) 17/21 -----P(AlB): 6/11 ------ P(BlA): 1/2 and the events to not be independent because (AnB) =/= P(A) x P(B)

I am not exactly sure if I'm right.

The second one is this one:

An unfair coin (Probability of 0.6 for heads) is flipped twice. Given that on at least one flip a head occurred, what is the probability that a head occurred on both flips.

So in my mind, this question is really simple, but the wording makes me believe that I am overlooking something and that it is a bit more complicated. am I supposed to do like an x chooses n / binomial formula thing.

Thank you in advance to anyone that can help me.

  • 2
    $\begingroup$ I do not see a problem with the answers to the first question, except you could use $\LaTeX$ to make it easier to read, for example $P(A \cap B)= \frac{6}{21}$ to give $P(A \cap B)= \frac{6}{21}$ . The second question is certainly shorter than the first, and is no harder. What did you think the answer was? $\endgroup$ – Henry Sep 16 '16 at 12:53
  • $\begingroup$ You could use the binomial formula, but it will be very cumbersome since there are only two flips. In this situation, treating every outcome by hand is easy. With 100 flips, binomial distribution is the way to go. $\endgroup$ – justt Sep 16 '16 at 12:56

Confirmations for die: $$P(A) = P\{2,4,6\} = (2+4+6)/21 = 12/21 = 4/7.$$ $$P(A \cap B) = P\{6\} = 6/21 = 2/7,$$ $$P(B|A) = P(A \cap B)/P(A) = 2/4 = 1/2.$$

Binomial computations for coin:

Random variable $X \sim Binom(2, .6)$ is the number of Heads in two tosses. Here there are $n = 2$ 'trials' with $P(\text{Success}) = P(\text{Heads}) = p = 0.6.$ Denote $q = 1-p.$ @justi is correct that using a binomial random variable could get cumbersome. But this is a good place to get started using the binomial distribution in a simple application. You want

$$P(X = 2 | X \ge 1) = \frac{P(X = 2,\,X\ge 1)}{P(X \ge 1)} = \frac{P(X=2)}{P(X\ge 1)} = \frac{p^2}{2pq + p^2}.$$

The comma in the numerator of the first fraction is commonly-used notation for $\cap$ between two expressions involving random variables. How do you justify the second equal sign? (If $A \subset B$, what is $A \cap B?$)

In R statistical software, where dbinom(k, 2, .6) computes $P(X = k) = {2 \choose k}p^k q^{n-k},$ for $k = 0, 1, 2.$

num = dbinom(2, 2, .6)
den = dbinom(1, 2, .6) + dbinom(2, 2, .6)
## 0.4285714

Note: It seems you are at the beginning of a very nice probability course. You may find it useful to install R on your Windows, Mac, or Linux computer, easy and free of charge from r-project.org. There is a lot to R and way too much if you try to learn everything. But just getting familiar with a dozen probability functions such as dbinom may save you some time doing tedious computations.


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