An integer is chosen at random from the integers 1, 2, 3,, 50. The probability that the chosen integer is a multiple of at least one of 4, 6, 7 is:

Given: An integer is chosen at random from the integers 1 to 50. Find: The probability that the chosen integer is a multiple of at least one of 4, 6, 7. Use the principle of inclusion-exclusion. Count the multiples of each number: Multiples of 4 = 12 since they are 4, 8, 12,, 48. Multiples of 6 = 8 since they are 6, 12, 18,, 48. Multiples of 7 = 7 since they are 7, 14, 21,, 49. Now count the common multiples of pairs: LCM(4,6)=12 4 common multiples up to 50 LCM(6,7)=42 1 common multiple up to 50 LCM(4,7)=28 1 common multiple up to 50 Also, LCM(4,6,7)=8450 so there is no common multiple of all three within the range. Apply inclusion-exclusion: aligned N(A B C) &= 12 + 8 + 7 - 4 - 1 - 1 + 0 &= 21 aligned So, 21 integers are multiples of at least one of 4, 6, 7. Therefore, the required probability is P = 2150 Hence, the correct option is B. Given: Let A be the event that the number is a multiple of 4, B be the event that it is a multiple of 6, and C be the event that it is a multiple of 7. Find: P(A B C). From the counts in 1 to 50, P(A)= 1250, P(B)= 850, P(C)= 750 Also, P(A B)= 450, P(B C)= 150, P(A C)= 150 and P(A B C)=0 Instead of listing all numbers, count directly using floor values: 504 = 12, 506 = 8, 507 = 7 Subtract pairwise overlaps: 5012 = 4, 5042 = 1, 5028 = 1 Add the triple overlap: 5084 = 0 So, 12+8+7-4-1-1+0=21 Thus the probability is 2150, so the correct option is B.

An integer is chosen at random from the integers 1, 2, 3,, 50. Th… — Solution | JEE Mathematics

MCQMediumJEE 2024Probability Basics

An integer is chosen at random from the integers $1, 2, 3, \ldots, 50$ . The probability that the chosen integer is a multiple of at least one of $4, 6, 7$ is:

A
$\frac{8}{25}$
B
$\frac{21}{50}$
C
$\frac{9}{50}$
D
$\frac{14}{25}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: An integer is chosen at random from the integers $1$ to $50$ .

Find: The probability that the chosen integer is a multiple of at least one of $4, 6, 7$ .

Use the principle of inclusion-exclusion.

Count the multiples of each number:

\text{Multiples of } 4 = 12

since they are $4, 8, 12, \ldots, 48$ .

\text{Multiples of } 6 = 8

since they are $6, 12, 18, \ldots, 48$ .

\text{Multiples of } 7 = 7

since they are $7, 14, 21, \ldots, 49$ .

Now count the common multiples of pairs:

\operatorname{LCM}(4,6)=12 \Rightarrow 4 \text{ common multiples up to } 50

\operatorname{LCM}(6,7)=42 \Rightarrow 1 \text{ common multiple up to } 50

\operatorname{LCM}(4,7)=28 \Rightarrow 1 \text{ common multiple up to } 50

Also,

\operatorname{LCM}(4,6,7)=84>50

so there is no common multiple of all three within the range.

Apply inclusion-exclusion:

\begin{aligned} N(A \cup B \cup C) &= 12 + 8 + 7 - 4 - 1 - 1 + 0 \\ &= 21 \end{aligned}

So, $21$ integers are multiples of at least one of $4, 6, 7$ .

Therefore, the required probability is

P = \frac{21}{50}

Hence, the correct option is B.

Event Probability Form

Given: Let $A$ be the event that the number is a multiple of $4$ , $B$ be the event that it is a multiple of $6$ , and $C$ be the event that it is a multiple of $7$ .

Find: $P(A \cup B \cup C)$ .

From the counts in $1$ to $50$ ,

P(A)=\frac{12}{50}, \quad P(B)=\frac{8}{50}, \quad P(C)=\frac{7}{50}

Also,

P(A \cap B)=\frac{4}{50}, \quad P(B \cap C)=\frac{1}{50}, \quad P(A \cap C)=\frac{1}{50}

and

P(A \cap B \cap C)=0

LCM Counting Shortcut

Instead of listing all numbers, count directly using floor values:

\left\lfloor \frac{50}{4} \right\rfloor = 12, \quad \left\lfloor \frac{50}{6} \right\rfloor = 8, \quad \left\lfloor \frac{50}{7} \right\rfloor = 7

Subtract pairwise overlaps:

\left\lfloor \frac{50}{12} \right\rfloor = 4, \quad \left\lfloor \frac{50}{42} \right\rfloor = 1, \quad \left\lfloor \frac{50}{28} \right\rfloor = 1

Add the triple overlap:

\left\lfloor \frac{50}{84} \right\rfloor = 0

So,

12+8+7-4-1-1+0=21

Thus the probability is $\frac{21}{50}$ , so the correct option is B.

Common mistakes

Adding the counts of multiples of $4, 6,$ and $7$ directly gives $12+8+7=27$ , which double-counts numbers like $12, 28,$ and $42$ . Use inclusion-exclusion and subtract overlaps.
Finding common multiples incorrectly by multiplying numbers instead of taking LCM leads to wrong intersections. For example, for $4$ and $6$ use $\operatorname{LCM}(4,6)=12$ , not $24$ .
Forgetting to check the triple intersection can cause an incomplete inclusion-exclusion calculation. Here $\operatorname{LCM}(4,6,7)=84$ , and since it exceeds $50$ , the triple overlap is $0$ .

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