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Zorluk: Çok zorConditional Probability

A box contains 12 cards numbered consecutively from 1 through 12. Two cards are selected at random from the box without replacement. Given that the sum of the numbers on the two selected cards is even, what is the probability that at least one of the selected cards has a prime number on it?

  1. 1930\frac{19}{30}Cevap
  2. B
    1966\frac{19}{66}
  3. C
    23\frac{2}{3}
  4. D
    715\frac{7}{15}
  5. E
    1522\frac{15}{22}

Cevap

The conditional probability that at least one selected card has a prime number on it, given that their sum is even, is 1930\frac{19}{30}.
The option offering 19/30 is correct. Given that the sum of the two cards is even, both cards must be even or both cards must be odd. Choosing 2 even cards from 6 available gives 15 pairs, and choosing 2 odd cards from 6 available gives 15 pairs, making 30 possible pairs in total for the restricted sample space. Among the even cards, 2 is prime while 4, 6, 8, 10, and 12 are non-prime (5 numbers). Among the odd cards, 3, 5, 7, and 11 are prime (4 numbers) while 1 and 9 are non-prime (2 numbers). The pairs containing no primes consist of 2 non-prime evens (10 pairs) and 2 non-prime odds (1 pair), giving 11 non-prime pairs. Subtracting from 30 yields 19 pairs with at least one prime. Thus, the conditional probability is 19/30.

Adım Adım Çözüm

1
Determine the restricted sample space (Condition B: Sum of two cards is even)
Total outcomes in Condition B = 30
The sum of two integers is even if both are even or both are odd. Among numbers 1 to 12, there are 6 even numbers ({2, 4, 6, 8, 10, 12}) and 6 odd numbers ({1, 3, 5, 7, 9, 11}). The number of ways to pick 2 even cards is (62)=15\binom{6}{2} = 15, and 2 odd cards is (62)=15\binom{6}{2} = 15. Total pairs with an even sum = 15+15=3015 + 15 = 30.
2
Categorize the numbers 1 through 12 by parity and primality
Prime evens = {2} (1 number); Non-prime evens = {4, 6, 8, 10, 12} (5 numbers); Prime odds = {3, 5, 7, 11} (4 numbers); Non-prime odds = {1, 9} (2 numbers)
Note that 1 is not a prime number, and 2 is the only even prime number.
3
Count the number of pairs in the restricted sample space with NO prime numbers
11 non-prime pairs
Pairs of two evens with no primes come from non-prime evens: (52)=10\binom{5}{2} = 10 pairs. Pairs of two odds with no primes come from non-prime odds: (22)=1\binom{2}{2} = 1 pair. Total non-prime pairs = 10+1=1110 + 1 = 11.
4
Calculate favorable outcomes (Event A ∩ B) and the conditional probability
P(A|B) = 19/30
Favorable pairs with at least one prime = 30−11=1930 - 11 = 19. Therefore, P(At least one prime∣Even sum)=1930P(\text{At least one prime} \mid \text{Even sum}) = \frac{19}{30}.

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Conditional Probability with Restricted Sample Space
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Conditional Probability

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