Answer to Question #171694 in Statistics and Probability for Melissa

Question

Two 4-sided dice are tossed 10 times and the sum of the “down” faces is noted. What is the probability that the sum of 4 occurs at most two times? Leave your answer as a percent rounded to one decimal place. Provide evidence of your work by typing out your full solution.

Answer 1

Solution:

Let $X$ be the random variable denoting a sum of 4 on two 4-sided dice.

Total number of possible outcomes$=4^2=16$

Favorable outcomes$=\{(1,3),(3,1),(2,2)\}$

Total number of favorable outcomes$=3$

$p=\dfrac 3{16}, q=1-\dfrac 3{16}=\dfrac {13}{16},n=10$

$X\sim \text{Binomial}(10,\dfrac 3{16})$

So, $P(X\le2)=P(X=0)+P(X=1)+P(X=2)$

$=^{10}C_0(\dfrac 3{16})^0(\dfrac {13}{16})^{10}+^{10}C_1(\dfrac 3{16})^1(\dfrac {13}{16})^{9}+^{10}C_2(\dfrac 3{16})^2(\dfrac {13}{16})^{8}$

$\approx0.7152=71.52\%$