1
$\begingroup$

I am reading "Schaum's Outline of Statistics". I understand the examples in the book, and as a result, I have produced my own probability question, but I am having difficulty in attempting to solve it.

Using Binomial Distribution:

A fair six-sided die is rolled once.

Calculate the probability of getting either a $1$ or a $4$.

This is what I have attempted:

$Success(P) = 2/6 = 0.\dot3$

$Failure(Q) = 4/6 = 0.\dot6$

$= (^1C_3) (0.\dot3)^1 (0.\dot6)^1-1$

$= 0.3\dot2$

This obviously is not correct. I am attempting to solve this question using Binomial Distribution. What am I doing incorrectly?

$\endgroup$
  • $\begingroup$ This is a duplicate of your prior question. $\endgroup$ – lulu Oct 24 '17 at 15:55
  • $\begingroup$ In truth, I have no guess what your question is. The answer appears to obviously be $\frac 26$. What on earth is the confusion? $\endgroup$ – lulu Oct 24 '17 at 15:56
  • $\begingroup$ I have voted to close this. If there is an underlying question here please try to ask it coherently. $\endgroup$ – lulu Oct 24 '17 at 15:57
  • $\begingroup$ Sorry for the confusion. Whilst the answer is 2/6, is it possible to solve this question using Binomial Distribution? $\endgroup$ – Scrub Oct 24 '17 at 15:57
  • $\begingroup$ If you know the answer is $\frac 26$ why do you say it is $.322$? Please delete this question and ask another, more carefully considered, one. $\endgroup$ – lulu Oct 24 '17 at 15:58
0
$\begingroup$

Using the binomial formula you still get the same answer:

You need $1$ success out of $1$ try, so:

$$P = {1 \choose 1} \cdot P(Success)^1 \cdot P(Failure)^0 = 1 \cdot \big(\frac{2}{6}\big)^1 \cdot \big(\frac{4}{6}\big)^0 = 1 \cdot \frac{2}{6} \cdot 1 = \frac{2}{6}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.