Question

case is going to toss three coins simultaneously. what is the probability that exactly two of the coins will land on heads? a. $\frac{1}{6}$ b. $\frac{1}{4}$ c. $\frac{1}{3}$ d. $\frac{3}{8}$ 4. an ordinary deck of cards has 4 suits: hearts, diamonds, clubs and spades. each suit has 13 different cards, so there are 52 cards in all. one card is picked at random from the deck. what is the probability that the cards suit is hearts or clubs? a. $\frac{1}{16}$ b. $\frac{2}{13}$ c. $\frac{1}{4}$ d. $\frac{1}{2}$

Explanation:

Step1: Calculate coin - toss probability

The total number of possible outcomes when tossing 3 coins is $2\times2\times2 = 2^3=8$ (since each coin has 2 possible outcomes: heads or tails).
We use the combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n = 3$ (number of coin - tosses) and $k = 2$ (number of heads we want).
$C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=\frac{3\times2!}{2!×1}=3$.
The probability of getting exactly 2 heads is $\frac{C(3,2)}{8}=\frac{3}{8}$.

Step2: Calculate card - pick probability

The number of hearts in a deck is 13 and the number of clubs is 13.
The probability of picking a heart is $P(H)=\frac{13}{52}$, and the probability of picking a club is $P(C)=\frac{13}{52}$.
Since the events of picking a heart and picking a club are mutually - exclusive, the probability of picking a heart or a club is $P(H\cup C)=P(H)+P(C)=\frac{13}{52}+\frac{13}{52}=\frac{13 + 13}{52}=\frac{26}{52}=\frac{1}{2}$.

Answer:

1. D. $\frac{3}{8}$
2. D. $\frac{1}{2}$