Question

sep 11.

1. a fair coin is flipped, and a 6-sided die is rolled. what is the probability of flipping a heads and rolling a 4? $\frac{1}{2} \times \frac{1}{3}$
2. a card is drawn at random from a standard deck of 52 cards. what is the probability of drawing a heart or a king?
3. a jar contains 3 red marbles, 5 blue marbles, and 2 green marbles. if one marble is randomly selected, what is the probability it is red or green?

Explanation:

Problem 1:

Step1: Probability of heads

A fair coin has 2 outcomes (heads, tails). Probability of heads: $\frac{1}{2}$.

Step2: Probability of rolling a 4

A 6 - sided die has 6 outcomes. Probability of rolling a 4: $\frac{1}{6}$. Wait, the original note has $\frac{1}{3}$, that's an error. Correctly, it's $\frac{1}{6}$.

Step3: Multiply the probabilities (independent events)

Since flipping a coin and rolling a die are independent, we multiply the probabilities: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$. But if we follow the incorrect $\frac{1}{3}$ in the note: $\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$. (But the correct die probability for 4 is $\frac{1}{6}$)

Step1: Define events

Let $A$ be the event of drawing a heart, $B$ be the event of drawing a king.

Step2: Calculate $n(A)$, $n(B)$, $n(A\cap B)$

In a standard deck, $n(A) = 13$ (hearts), $n(B)=4$ (kings), $n(A\cap B) = 1$ (king of hearts). Total cards $n(S)=52$.

Step3: Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$P(A)=\frac{13}{52}$, $P(B)=\frac{4}{52}$, $P(A\cap B)=\frac{1}{52}$.
So $P(A\cup B)=\frac{13 + 4- 1}{52}=\frac{16}{52}=\frac{4}{13}$.

Step1: Calculate total number of marbles

Total marbles $= 3+5 + 2=10$.

Step2: Define events

Let $A$ be the event of selecting a red marble, $B$ be the event of selecting a green marble. $A$ and $B$ are mutually exclusive (a marble can't be both red and green).

Step3: Calculate $n(A)$ and $n(B)$

$n(A) = 3$, $n(B)=2$.

Step4: Use the formula for mutually exclusive events $P(A\cup B)=P(A)+P(B)$

$P(A)=\frac{3}{10}$, $P(B)=\frac{2}{10}$. So $P(A\cup B)=\frac{3 + 2}{10}=\frac{5}{10}=\frac{1}{2}$.

Answer:

If we use the correct die probability: $\frac{1}{12}$; If we use the note's incorrect $\frac{1}{3}$: $\frac{1}{6}$

Problem 2: