Study questions platform-wide or filter by specific tests with correct answers revealed.
Two cards are drawn one after another with replacement from a well-shuffled standard deck of 52 cards. Let $X$ be the number of aces obtained. Find $P(X=1)+P(X=2)$.
$p=P(\text{ace})=\dfrac{4}{52}=\dfrac{1}{13}$, $q=\dfrac{12}{13}$. Binomial with $n=2$.
$P(X=1)=\binom{2}{1}\cdot\dfrac{1}{13}\cdot\dfrac{12}{13}=\dfrac{24}{169}$
$P(X=2)=\binom{2}{2}\cdot\left(\dfrac{1}{13}\right)^2=\dfrac{1}{169}$
$P(X=1)+P(X=2)=\dfrac{24}{169}+\dfrac{1}{169}=\dfrac{25}{169}$
Given that $|-\overline{3}| = -m$, compare:
Column A: $m$
Column B: 3
First, evaluate the left side: $|-\overline{3}| = |-3| = 3$
So we have: $3 = -m$
Solving for $m$: $m = -3$
Now compare:
- Column A: $m = -3$
- Column B: $3$
Since $-3 < 3$, Column B is greater.
Sign in to join the conversation and share your thoughts.
Log In to Comment