14. Exercises (Examples)

1. Two cards drawn from a standard deck without replacement: Probability both are hearts?

To determine the probability of drawing two hearts from a standard deck of 52 cards without replacement, we follow these steps:

1. Identify the Sample Space: A standard deck has 52 cards with 13 of them being hearts.
2. Calculate the Probability of the First Card: The chance of drawing a heart first is 13 out of 52, or 13/52.
3. Calculate the Probability of the Second Card: After drawing one heart, there are now 51 cards left, with only 12 hearts remaining. Thus, the probability of drawing a second heart is 12 out of 51, or 12/51.
4. Combine the Probabilities: The overall probability is obtained by multiplying the probabilities of both events together: (13/52) * (12/51). Performing this multiplication gives the final probability of both cards being hearts.

Imagine you have a bag of marbles—13 red ones (representing hearts) and 39 blue ones (non-hearts). If you draw one marble without looking, the chances of it being red are 13 out of 52. After removing a red marble, you now have 12 reds left in a bag of 51. Drawing a second red marble gets trickier since there’s now a smaller pool to choose from.

2. A biased coin with P(heads)=0.6 tossed 3 times: Distribution of number of heads.

To calculate the distribution of heads when tossing a biased coin three times, we can perform the following steps:

1. Understanding the Probability: The probability of getting heads on any single toss is 0.6, while the probability of getting tails is 0.4 (1 - 0.6).
2. Identifying Outcomes: There are 8 different possible outcomes when tossing the coin 3 times (e.g., HHH, HHT, HTH, HTT, THH, THT, TTH, TTT).
3. Calculating Individual Probabilities: Using the binomial formula, we can calculate the probabilities of getting 0, 1, 2, or 3 heads:
   • P(0 heads) = (0.4)^3 = 0.064
   • P(1 head) = 3 * (0.6)^1 * (0.4)^2 = 3 * 0.6 * 0.16 = 0.288
   • P(2 heads) = 3 * (0.6)^2 * (0.4)^1 = 3 * 0.36 * 0.4 = 0.432
   • P(3 heads) = (0.6)^3 = 0.216
4. Summarizing: The distribution of heads can thus be summarized as:
   • 0 heads: P = 0.064
   • 1 head: P = 0.288
   • 2 heads: P = 0.432
   • 3 heads: P = 0.216.

Think of a vending machine where every time you want to buy a drink, it sometimes gives you a soda (heads) and sometimes just gives you your money back (tails). If it favors giving you a drink 60% of the time, you can figure out over 3 tries how many sodas you are likely to get and how often you’ll just get your money back.

3. Traffic light probability problem: Given it’s red with prob. 0.4, what is P(green | not red)?

To solve this traffic light problem, we need to use conditional probability:

1. Understand Given Probabilities: The probability that the light is red (P(red)) is 0.4. Therefore, the probability that it’s not red (P(not red)) is 0.6 (1 - 0.4).
2. Calculate P(green): Since the light can only be red, green, or yellow, and assuming that yellow is not considered in this question, we can derive that P(green) = 1 - P(red) - P(yellow). Without a specific P(yellow), we assume yellow has a negligible impact in this scenario. Thus, P(green) = P(not red).
3. Final Calculation: Therefore, P(green | not red) = P(green) / P(not red) = P(green) / 0.6. If we assume equal likelihood for green and yellow, we find that P(green) = 0.5. Hence, P(green | not red) = 0.5/0.6, simplifying to 5/6, or approximately 0.83.

Imagine you're waiting to cross the road, and you know the light is red 40% of the time. If you see that it’s not red, you might wonder how likely it is to be green instead of yellow. Realizing that every time it isn’t red, you’d have a high chance of seeing the green light—like skipping through a crowded area where less popular attractions attract fewer people but the best ones are still the most crowded.

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