1) A basket contains 3 red and 2 yellow apples. Two apples are chosen at random. Find the following probabilities:
a) P(one red, one yellow) b) P(at least one red)
2) A card is drawn from a standard deck. Find the following probabilities:
a) P(a jack or a king) b) P(a jack or a spade)
3) Mrs. Rossetti is flying from San Francisco to New York. On her way to the San Francisco Airport she encounters heavy traffic and determines that there is a 20% chance that she will be late to the airport and will miss her flight. Even if she makes her flight, there is a 10% chance that she will miss her connecting flight at Chicago. What is the probability that she will make it to New York as scheduled?
4) At a college, twenty percent of the students take history, thirty percent take math, and ten percent take both. What percent of the students take at least one of these two courses?
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1) To find the probabilities in this scenario, we can use the concept of combinations and the formula for probability.
a) P(one red, one yellow):
First, let's determine the total number of possible combinations when choosing two apples from the basket. Since there are 5 apples in total, we can calculate this using the combination formula: C(5, 2) = 5! / (2! * (5-2)!) = 10.
Now, let's determine the number of combinations that have one red and one yellow apple. Since there are 3 red apples and 2 yellow apples, we can calculate this using the combination formula again: C(3, 1) * C(2, 1) = (3! / (1! * (3-1)!)) * (2! / (1! * (2-1)!)) = 3 * 2 = 6.
Finally, we can find the probability by dividing the number of favorable outcomes (one red, one yellow) by the total number of possible outcomes: P(one red, one yellow) = 6/10 = 0.6.
b) P(at least one red):
To find the probability of having at least one red apple, we can consider the complementary event (i.e., the event of not having any red apples) and subtract it from 1.
The probability of not having any red apples is equal to the probability of choosing both apples as yellow: P(no red) = C(2, 2) / C(5, 2) = 1/10 = 0.1.
Therefore, the probability of having at least one red apple is equal to 1 minus the probability of not having any red apples: P(at least one red) = 1 - P(no red) = 1 - 0.1 = 0.9.
2) To find the probabilities in this scenario, we also need to consider combinations and use the formula for probability.
a) P(a jack or a king):
First, let's determine the total number of possible outcomes when drawing one card from a standard deck, which consists of 52 cards.
The number of jacks is 4, and the number of kings is also 4. However, we should not count the card that is both a jack and a king (i.e., the jack of hearts) twice.
Therefore, the number of favorable outcomes is 4 + 4 - 1 = 7 (since we subtract 1 for the overlapping card).
Finally, we can find the probability by dividing the number of favorable outcomes by the total number of possible outcomes: P(a jack or a king) = 7/52 = 0.1346 (approximately).
b) P(a jack or a spade):
To find the probability of drawing a jack or a spade, we can consider the total number of jacks and the total number of spades separately.
The number of jacks is 4, and the total number of spades is 13.
Again, we need to subtract the overlapping card (the jack of spades) from the total count.
Therefore, the number of favorable outcomes is 4 + 13 - 1 = 16.
Finally, we can find the probability by dividing the number of favorable outcomes by the total number of possible outcomes: P(a jack or a spade) = 16/52 = 0.3077 (approximately).
3) To determine the probability that Mrs. Rossetti will make it to New York as scheduled, we can use the concept of conditional probability.
Let's consider the possible scenarios:
1) She is late for her flight and misses it,
2) She is on time for her flight but misses her connecting flight,
3) She is on time for both flights.
a) The probability of being late for her flight is given as 20%.
b) The probability of missing her connecting flight, even if she makes her first flight, is 10%.
To find the probability that she will make it to New York as scheduled, we need to calculate the probability of scenario 3 (being on time for both flights) by considering the complement of scenarios 1 and 2.
The probability of scenario 1 is 20%, so the complement (not being late for her flight) is 1 - 20% = 80%.
The probability of scenario 2 is 10%, so the complement (not missing her connecting flight) is 1 - 10% = 90%.
Now, we can calculate the probability of scenario 3 (being on time for both flights) by multiplying the probability of not being late and the probability of not missing the connecting flight:
P(scenario 3) = (80%) * (90%) = 0.8 * 0.9 = 0.72.
Therefore, the probability that Mrs. Rossetti will make it to New York as scheduled is 72%.
4) To find the percentage of students who take at least one of the two courses (history and math), we need to consider the intersection and unions of events.
Let's denote:
H = event that a student takes history
M = event that a student takes math
We are given:
P(H) = 20% = 0.20
P(M) = 30% = 0.30
P(H ∩ M) = 10% = 0.10
To find the probability that a student takes at least one of the two courses, we can use the principle of inclusion-exclusion. The formula is:
P(H ∪ M) = P(H) + P(M) - P(H ∩ M)
Substituting the given values:
P(H ∪ M) = 0.20 + 0.30 - 0.10 = 0.40
Therefore, 40% of the students take at least one of the two courses (history or math).