Question

o option a: draw through a deck of shuffled cards without replacement, defining \success\ as \the card drawn is red (either a heart or a diamond).\ - determine the theoretical probability of success for your chosen option. (if selecting option c, you may have to do a web search for this information) - run your experiment twenty times. record your number of successes. - compute the empirical probability of success using your results. - using your theoretical probability, compute the probability of observing two successes in a row. - using your empirical probability, compute the probability of observing two successes in a row. - determine two conditional probabilities for your option: o option a: suppose that the first ten cards were revealed to all be black (either a spade or a club). what is the probability that the next card will be red? repeat this question but with the first ten cards all being red. - make a conclusion. o when you repeated the experiment 20 times, should we consider each trial dependent or independent from each other?

Explanation:

Step1: Calculate theoretical probability of success

A standard deck has 52 cards, 26 of which are red. So the theoretical probability $P(\text{success})$ is $\frac{26}{52}=\frac{1}{2}$.

Step2: Empirical probability after 20 - time experiment

Since the number of successes in the 20 - time experiment is not given, assume the number of successes is $n$. Then the empirical probability $P_{empirical}=\frac{n}{20}$.

Step3: Probability of two theoretical successes in a row

Since the theoretical probability of success in one draw is $\frac{1}{2}$, and the draws are independent (in theory), the probability of two successes in a row is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.

Step4: Probability of two empirical successes in a row

If the empirical probability is $P_{empirical}=\frac{n}{20}$, the probability of two empirical successes in a row is $(\frac{n}{20})\times(\frac{n}{20})=\frac{n^{2}}{400}$.

Step5: Conditional probability when first 10 cards are black

After 10 black cards are drawn, there are $52 - 10=42$ cards left, and 26 of them are red. So the probability that the next card is red is $\frac{26}{42}=\frac{13}{21}$.

Step6: Conditional probability when first 10 cards are red

After 10 red cards are drawn, there are $52 - 10 = 42$ cards left, and $26-10 = 16$ of them are red. So the probability that the next card is red is $\frac{16}{42}=\frac{8}{21}$.

Step7: Conclusion about independence

When repeating the experiment 20 times, in the case of drawing cards without replacement, the trials are dependent. Because the outcome of one draw affects the composition of the deck for the next draw.

Answer:

Theoretical probability of success: $\frac{1}{2}$
Empirical probability of success (assuming $n$ successes in 20 trials): $\frac{n}{20}$
Probability of two theoretical successes in a row: $\frac{1}{4}$
Probability of two empirical successes in a row (assuming $n$ successes in 20 trials): $\frac{n^{2}}{400}$
Conditional probability (first 10 cards black): $\frac{13}{21}$
Conditional probability (first 10 cards red): $\frac{8}{21}$
Trials are dependent.