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?

  • $\begingroup$ 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

Using the binomial formula you still get the same answer:

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

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


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