diff --git a/06_StatisticalInference/homework/hw1.Rmd b/06_StatisticalInference/homework/hw1.Rmd
index 7f31c63a0..f5476f5c7 100644
--- a/06_StatisticalInference/homework/hw1.Rmd
+++ b/06_StatisticalInference/homework/hw1.Rmd
@@ -40,22 +40,22 @@ Creating Data Products
 
 --- &radio
 
-Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 15% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 6% while that the mother contracted the disease is 5%. What is the probability that both contracted influenza expressed as a whole number percentage?
+Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 15% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 10% while that the mother contracted the disease is 9%. What is the probability that both contracted influenza expressed as a whole number percentage?
 
 1. 15%
-2. 6%
-3. 5%
-4. _2%_
+2. 10%
+3. 9%
+4. _4%_
 
 *** .hint
-$A = Father$, $P(A) = .06$, $B = Mother$, $P(B) = .05$ 
+$A = Father$, $P(A) = .10$, $B = Mother$, $P(B) = .09$ 
 $P(A\cup B) = .15$, 
 
 *** .explanation
-$P(A\cup B) = P(A) + P(B) - 2 P(AB)$ thus
-$$.15 = .06 + .05 - 2 P(AB)$$
+$P(A\cup B) = P(A) + P(B) - P(AB)$ thus
+$$.15 = .10 + .09 - P(AB)$$
 ```{r}
-(0.15 - .06 - .05) / 2
+.10 + .09 - .15
 ```
 
 ---  &radio
diff --git a/06_StatisticalInference/homework/hw2.Rmd b/06_StatisticalInference/homework/hw2.Rmd
index 3a568425c..b38eab299 100644
--- a/06_StatisticalInference/homework/hw2.Rmd
+++ b/06_StatisticalInference/homework/hw2.Rmd
@@ -157,10 +157,10 @@ Let $p=.5$ and $X$ be binomial
 
 *** .explanation
 
-<span class="answer">`r round(pbinom(4, prob = .5, size = 6, lower.tail = TRUE) * 100, 1)`</span>
+<span class="answer">`r round(pbinom(4, prob = .5, size = 6, lower.tail = FALSE) * 100, 1)`</span>
 
 ```{r}
-round(pbinom(4, prob = .5, size = 6, lower.tail = TRUE) * 100, 1)
+round(pbinom(4, prob = .5, size = 6, lower.tail = FALSE) * 100, 1)
 ```
 
 --- &multitext
@@ -210,9 +210,9 @@ If you roll ten standard dice, take their average, then repeat this process over
 $$Var(\bar X) = \sigma^2 /n$$
 
 *** .explanation
-The answer will be <span class="answer">`r round( mean(1 : 6 - 3.5) ^2 / 100, 3)`</span> 
-since the variance of the sampling distribution of the mean is $\sigma^2/12$
-and the variance of a die roll is 
+The answer will be <span class="answer">`r round( mean( (1 : 6 - 3.5) ^2) / 10, 3)`</span> 
+since the variance of the sampling distribution of the mean is $\sigma^2/10$
+where $\sigma^2$ is the variance of a single die roll, which is 
 
 ```{r}
 mean((1 : 6 - 3.5)^2)