Question

match the desired compound events with their corresponding probabilities. roll a number cube and get a 4 and 5. flip a coin to get tails up and roll a number cube to get an even number. draw an ace and then a 7 from a deck of cards with replacement. flip a coin 3 times and get heads every time. 1/8 1/4 1/169 1/36

Explanation:

Step1: Probability of getting 4 and 5 on a number - cube

A number - cube has 6 faces numbered from 1 to 6. It is impossible to get a 4 and 5 simultaneously on a single roll. So the probability is 0. But among the given options, we assume there is a mis - statement and we consider the probability of getting a 4 or 5. The probability of getting a 4 on a roll of a number cube is $P(4)=\frac{1}{6}$, and the probability of getting a 5 on a roll of a number cube is $P(5)=\frac{1}{6}$. Since these are mutually exclusive events, $P(4\ or\ 5)=\frac{1 + 1}{6}=\frac{1}{3}$ (not in the options, but if we consider the problem as getting 4 and then 5 in two rolls, for independent events, $P(4)\times P(5)=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$).

Step2: Probability of getting tails on a coin - flip and an even number on a number - cube

The probability of getting tails on a coin - flip is $P(T)=\frac{1}{2}$. The probability of getting an even number (2, 4, or 6) on a number - cube is $P(E)=\frac{3}{6}=\frac{1}{2}$. Since these are independent events, by the multiplication rule of independent events $P(T\ and\ E)=P(T)\times P(E)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.

Step3: Probability of drawing an ace and then a 7 from a deck of cards with replacement

A standard deck of cards has 52 cards. The probability of drawing an ace is $P(A)=\frac{4}{52}=\frac{1}{13}$. Since we replace the card, the probability of drawing a 7 is $P(7)=\frac{4}{52}=\frac{1}{13}$. For independent events, $P(A\ and\ 7)=P(A)\times P(7)=\frac{1}{13}\times\frac{1}{13}=\frac{1}{169}$.

Step4: Probability of flipping a coin 3 times and getting heads every time

The probability of getting heads on a single coin - flip is $P(H)=\frac{1}{2}$. Since the coin - flips are independent events, the probability of getting heads 3 times in a row is $P(H\ and\ H\ and\ H)=P(H)\times P(H)\times P(H)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$.

Answer:

Roll a number cube and get a 4 and 5 $\to$ $\frac{1}{36}$
Flip a coin to get tails up and roll a number cube to get an even number $\to$ $\frac{1}{4}$
Draw an ace and then a 7 from a deck of cards with replacement $\to$ $\frac{1}{169}$
Flip a coin 3 times and get heads every time $\to$ $\frac{1}{8}$