Supplemental Module – Binomial Distribution: Background You’ll Need 2
Compute probabilities that involve exponents
How Many 1’s?
When events [latex]A[/latex] and [latex]B[/latex] are independent, then [latex]P(A \text{ and } B) = P(A) \cdot P(B)[/latex]
When using this rule, there are situations in which the probabilities of some events are the same, and that’s when exponents come into play.
Exponent Review
[latex]p^{x}[/latex] means that [latex]p[/latex] is multiplied by itself [latex]x[/latex] times. For example: [latex](0.5)^{2} = (0.5)\cdot (0.5) = 0.25, \mbox{ and } (0.5)^{3} = (0.5)\cdot (0.5)\cdot (0.5) = 0.125[/latex].
Multiplication is commutative, which means that the order we multiply numbers in does not matter when multiplication is the only operation present. For example: [latex](0.25)\cdot (0.5)\cdot (0.25) = (0.25)\cdot (0.25)\cdot (0.5) = (0.25)^{2}(0.5)[/latex]
Suppose we are rolling a fair, [latex]6[/latex]-sided die [latex]4[/latex] times.Calculate the probability that the first roll is not a [latex]6[/latex], the second roll is a [latex]6[/latex], and the final two rolls are also not [latex]6[/latex].
The probability that the first roll is not a [latex]6[/latex], the second roll is a [latex]6[/latex], and the final two rolls are also not [latex]6[/latex] is:
[latex]P([/latex]not [latex]6[/latex] AND [latex]6[/latex] AND not [latex]6[/latex] AND not [latex]6)[/latex] = [latex]P(\text{not} 6)[/latex] [latex]\cdot[/latex] [latex]P(6)[/latex] [latex]\cdot[/latex] [latex]P(\text{not} 6)[/latex] [latex]\cdot[/latex] [latex]P(\text{not} 6)[/latex]