A number is selected at random from a list of the first 10 positive in

Question

A number is selected at random from a list of the first 10 positive integers. What is the probability that the number selected will be odd or divisible by 3?

A. \(\frac{3}{20}\)

B. \(\frac{1}{2}\)

C. \(\frac{13}{20}\)

D. \(\frac{3}{5}\)

E. \(\frac{4}{5}\)

P(odd) = 5/10 = 1/2
P( div. By 3 ) = 3/10
Req probability = 1/2 +3/10 = 4/5

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Sanyam_Maheshwari · Answer

To find the probability that the number selected will be odd or divisible by 3, we first need to determine the count of numbers that satisfy either condition and then divide it by the total number of possible outcomes.

The first 10 positive integers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Odd numbers: 1, 3, 5, 7, 9 (5 numbers)
Numbers divisible by 3: 3, 6, 9 (3 numbers)
Now, we need to find the numbers that satisfy either condition without double-counting. The numbers that are both odd and divisible by 3 are 3 and 9.

So, the total count of numbers that are either odd or divisible by 3 is: 5 (odd) + 3 (divisible by 3) - 2 (both) = 6.

The total number of possible outcomes is 10.

Now, we can calculate the probability:

Probability= Number of Favorable Outcomes / Total Number of Possible Outcomes
​
Probability = 6/10 = 0.6 =3/5

Therefore, the probability that the number selected will be odd or divisible by 3 is 3/5

Bunuel · Answer

Note that there is an overlap between odd numbers and multiples of 3. Specifically, 3 and 9 are both odd and multiples of 3, so the probability of this overlap should be subtracted from your result. This leads to 4/5 - 2/10 = 3/5.

Answer: D.

sayan640 · Answer

Total odd numbers = 5
Total numbers divisible by 3 = 3 (3,6 and 9)
Total no. of numbers which are odd and divisible by 3 = 2 (3 and 9 only )
Total no of numbers which are odd or divisible by 3 = 5 + 3 -2 = 6
req prob = 6/10 = 3/5

jaknah · Answer

Solving it in the most simplistic way would be: 
P(A or B) = P(A) + P (B) - P ( A intersection B) 
Hence, P(A) = 1,3,5,7,9 / 10 = 5/10
P(B) = 3,6, 9 / 10 i.e = Total of 3 numbers divisible / 10 = 3/10
P(A intersection B ) = Only 2 numbers are both odd and divisible by 3. Hence, 2/10
Now, solve for the equation, P(A or B) = P(A) + P (B) - P ( A intersection B) 
P(A or B) = 5/10 + 3/10 - 2/10 = 6/10 or 3/5

GMATinsight · Answer

first 10 positive integers = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = 10 numbers

Odd or divisible by 3 among them are {1, 3, 5, 6, 7, 9} = 6 numbers

Probability = 6/10 = 3/5

Answer: Option D

Related HARD Qn Video: 
  https://www.youtube.com/watch?v=xWR1VpyCKP0

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