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Zorluk: Çok zorBasic Probability and Counting Methods

A high school debate club consists of 15 members: 6 freshmen, 5 sophomores, and 4 juniors. If a committee of 3 members is to be selected at random from the club, what is the probability that the committee will contain at least 1 freshman and at least 1 sophomore?

  1. A
    2491\frac{24}{91}
  2. B
    247455\frac{247}{455}
  3. 5191\frac{51}{91}Cevap
  4. D
    251455\frac{251}{455}
  5. E
    67\frac{6}{7}

Cevap

The correct probability is 5191\frac{51}{91}.
To find the probability that a randomly chosen 3-member committee from a club of 15 members (6 freshmen, 5 sophomores, 4 juniors) has at least 1 freshman and at least 1 sophomore, we can find the total number of possible committees and subtract the number of unfavorable committees. The total number of committees is (153)=455\binom{15}{3} = 455. The unfavorable committees are those with no freshmen (chosen from the 9 sophomores and juniors: (93)=84\binom{9}{3} = 84) or no sophomores (chosen from the 10 freshmen and juniors: (103)=120\binom{10}{3} = 120). Because these two groups overlap when only juniors are chosen ((43)=4\binom{4}{3} = 4), the total number of unfavorable committees is 84+120−4=20084 + 120 - 4 = 200 by the Principle of Inclusion-Exclusion. Thus, there are 455−200=255455 - 200 = 255 favorable committees, yielding a probability of 255455=5191\frac{255}{455} = \frac{51}{91}.

Adım Adım Çözüm

1
Calculate the total number of possible 3-member committees that can be formed from the 15 club members.
(153)=15×14×133×2×1=455\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 total committees.
To find the probability, we first need to determine the size of the entire sample space (the total number of possible outcomes).
2
Determine the number of committees that do not meet the requirement of containing at least one freshman and at least one sophomore. This occurs if a committee has no freshmen, no sophomores, or neither.
Let AA be the event of choosing a committee with no freshmen (only sophomores and juniors): (93)=84\binom{9}{3} = 84 ways. Let BB be the event of choosing a committee with no sophomores (only freshmen and juniors): (103)=120\binom{10}{3} = 120 ways. Let A∩BA \cap B be the event of choosing a committee with neither freshmen nor sophomores (only juniors): (43)=4\binom{4}{3} = 4 ways.
It is easier to count the complement (the unfavorable outcomes) and subtract it from the total.
3
Use the Principle of Inclusion-Exclusion to find the total number of unfavorable committees (no freshmen or no sophomores).
∣A∪B∣=∣A∣+∣B∣−∣A∩B∣=84+120−4=200|A \cup B| = |A| + |B| - |A \cap B| = 84 + 120 - 4 = 200 unfavorable committees.
Since the events of having no freshmen and having no sophomores overlap when only juniors are selected, we must subtract the intersection to avoid double-counting.
4
Subtract the number of unfavorable committees from the total number of committees to find the number of favorable committees, and then compute the probability.
Favorable committees: 455−200=255455 - 200 = 255 ways. Probability: 255455=5191\frac{255}{455} = \frac{51}{91}.
Subtracting the complement from the total gives the number of valid outcomes, and dividing this by the total outcomes yields the desired probability.

Anahtar Kavram

Using combinations and the Principle of Inclusion-Exclusion to calculate probabilities of compound events.
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Konular

Basic Probability and Counting Methods

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