You can also use the the "binomial probability" form.
=C\large\begin{pmatrix}n\\x\end{pmatrix}p^xq^{ n-x})
Where
!x!})
Where x is the number of successes we want, n is the number trials, p is the chance of success, and q is the chance of failure.
For this question we have
P(1)+P(2)+P(3) (Because we want the odds of 1 heart, 2 hearts, and 3 hearts)
x = 1..2...3, n=3, p=0.25, q=0.75
+P(2)+P(3) = \frac{3!}{(3-1)!1!}(0.25)^1(0.75)^{3-1}+\frac{3!}{(3-2)!2!}(0.25)^2(0.75)^{3-2}+\frac{3!}{(3-3)!3!}(0.25)^3(0.75)^{3-3})
simplified we have
+P(2)+P(3) = 3(0.25)(0.75)^{2}+3(0.25)^2(0.75)+(0.25)^3)
And this also ends up being about 57.8 percent
Linear Mode
