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Zorluk: Çok zorBasic Single-Event Probability

In a game, a fair wheel lands on a single integer from 11 to 100100, inclusive, with each integer being equally likely to be selected. What is the probability that the integer selected is either a prime number or leaves a remainder of 11 when divided by 55, but not both?

  1. 720\frac{7}{20}Cevap
  2. B
    1750\frac{17}{50}
  3. C
    25\frac{2}{5}
  4. D
    920\frac{9}{20}
  5. E
    120\frac{1}{20}

Cevap

The correct answer is 7/20.
The total number of possible outcomes is 100. There are 25 prime numbers in the range 1 to 100, and 20 numbers that leave a remainder of 1 when divided by 5. Exactly 5 numbers (11, 31, 41, 61, 71) meet both conditions. To find the numbers that meet either condition but NOT both, subtract the 5 overlapping numbers from each category: 20 numbers are prime only, and 15 numbers leave a remainder of 1 only. The number of favorable outcomes is 20 + 15 = 35. Therefore, the single-event probability is 35/100 = 7/20.

Adım Adım Çözüm

1
Determine the total sample space size.
The total number of possible outcomes is N=100N = 100.
Each integer from 1 to 100 inclusive is equally likely.
2
Identify and count all prime numbers in the set {1,2,,100}\{1, 2, \dots, 100\}.
There are 25 prime numbers: {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}.
Note that 1 is not a prime number.
3
Identify and count all integers in {1,2,,100}\{1, 2, \dots, 100\} that leave a remainder of 1 when divided by 5.
There are 20 such integers: {1,6,11,16,21,26,31,36,41,46,51,56,61,66,71,76,81,86,91,96}\{1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, 91, 96\}.
These integers take the form 5k+15k + 1 for integers k0k \ge 0 where 15k+11001 \le 5k + 1 \le 100.
4
Find the intersection of both sets (numbers that are prime AND leave a remainder of 1 when divided by 5).
The 5 numbers satisfying both conditions are {11,31,41,61,71}\{11, 31, 41, 61, 71\}.
Any integer leaving a remainder of 1 when divided by 5 must end in 1 or 6. No prime ends in 6. The primes ending in 1 in this range are 11, 31, 41, 61, and 71.
5
Calculate the number of favorable outcomes satisfying 'either condition, but not both' (symmetric difference).
Favorable outcomes = (255)+(205)=20+15=35(25 - 5) + (20 - 5) = 20 + 15 = 35.
We must count numbers that are only prime (20) plus numbers that are only remainder 1 (15), excluding the 5 overlapping numbers.
6
Calculate the single-event probability.
P=35100=720P = \frac{35}{100} = \frac{7}{20}.
Divide favorable outcomes by total sample space size.

Anahtar Kavram

Basic single-event probability with set operations and number properties (prime numbers, remainders).
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