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Zorluk: ZorBasic Single-Event Probability

A container holds 60 cards, each marked with a distinct integer from 1 to 60, inclusive. If one card is selected at random, the probability that the integer on the selected card is divisible by 5, but divisible by neither 4 nor 6, can be expressed as a simplified fraction ab\frac{a}{b}, where aa and bb are positive integers with no common factors greater than 1. What is the value of a+ba + b?

Cevap: 17

Cevap

The value of a+ba + b is 17.
There are 60 equally likely outcomes. Out of 12 multiples of 5, exactly 4 integers (20, 30, 40, and 60) are divisible by 4 or 6. Thus, 12 - 4 = 8 integers satisfy the condition of being divisible by 5 but neither 4 nor 6. The single-event probability is 8/60 = 2/15. In simplest form, a = 2 and b = 15, yielding a + b = 17.

Adım Adım Çözüm

1
Determine total number of possible outcomes.
Total outcomes N=60N = 60.
Each of the 60 cards is equally likely to be selected.
2
Identify all multiples of 5 in the set {1,2,…,60}\{1, 2, \dots, 60\}.
Multiples of 5: {5,10,15,20,25,30,35,40,45,50,55,60}\{5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60\}, containing 605=12\frac{60}{5} = 12 integers.
These represent all potential candidates before applying exclusion criteria.
3
Identify multiples of 5 that are also divisible by 4 or by 6.
Multiples of 5 divisible by 4: LCM(5,4)=20→{20,40,60}\text{LCM}(5,4) = 20 \rightarrow \{20, 40, 60\} (3 integers). Multiples of 5 divisible by 6: LCM(5,6)=30→{30,60}\text{LCM}(5,6) = 30 \rightarrow \{30, 60\} (2 integers). Overlap (divisible by both 4 and 6): LCM(5,4,6)=60→{60}\text{LCM}(5,4,6) = 60 \rightarrow \{60\} (1 integer).
An integer is divisible by both 5 and 4 if it is a multiple of 20, and by both 5 and 6 if it is a multiple of 30.
4
Apply the inclusion-exclusion principle to find the count of invalid multiples of 5.
Invalid multiples = 3+2−1=43 + 2 - 1 = 4 integers ({20,30,40,60}\{20, 30, 40, 60\}). Favorable outcomes = 12−4=812 - 4 = 8 integers ({5,10,15,25,35,45,50,55}\{5, 10, 15, 25, 35, 45, 50, 55\}).
The integer 60 is divisible by both 4 and 6 and must not be double-subtracted.
5
Calculate the probability, express it as a simplified fraction ab\frac{a}{b}, and compute a+ba + b.
Probability P=860=215P = \frac{8}{60} = \frac{2}{15}. Thus, a=2a = 2, b=15b = 15, and a+b=2+15=17a + b = 2 + 15 = 17.
Dividing the numerator and denominator by 4 simplifies 860\frac{8}{60} to 215\frac{2}{15}, where gcd(2,15)=1\text{gcd}(2, 15) = 1.

Anahtar Kavram

Basic Single-Event Probability with Multiple Divisibility Constraints
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Konular

Basic Single-Event Probability

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