Question: A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
- 1/4
- 3/8
- 1/2
- 5/8
- 3/4
“A box contains 100 balls, numbered from 1 to 100”- is a topic covered under the GMAT Quantitative reasoning section of GMAT. GMAT quant measures the efficiency of the students to utilise analytical and logical reasoning to solve calculative problems. The candidate must pick the right option among the five given options by considering it mathematically. In order to solve GMAT Problem Solving questions, a student must possess better qualitative knowledge. The calculative mathematical problems of the GMAT Quant topic in the problem-solving part should be cracked with good mathematical understanding. The candidates can practice more questions from the book “The Official Guide for GMAT Quantitative Review” to enhance their knowledge.
Solution and Explanation:
Approach Solution 1:
The problem statement informs that:
Given:
- A box contains 100 balls from 1 to 100.
- 3 balls are picked randomly from the box.
Find Out:
- With the replacement of the box find the probability that the sum of three numbers on the balls selected from the box will be odd.
Since three balls are selected randomly from the box, it may give an odd sum or even sum of the three balls.
The odd sum of balls may be: OEE, EOE EEO, OOO
The even sum of balls may be: EEE, EOO, OEO, OOE.
It does not matter whether it is with or without a replacement case, the probability of an odd sum and an even sum will be equal. This is because there is an equal number of odd and even balls.
Since the above events depict all probable outcomes of selecting 3 three balls and are mutually exclusive, then their aggregate must be 1.
P(odd sum) = P (even sum) = ½
Since we need to find the probability of the sum of three balls,
Therefore, P (odd sum) = P (OEE) + P(EOE) + P(EEO) + P(OOO)
= ⅛ +⅛ +⅛ +⅛
=½
Correct Answer: (C)
Approach Solution 2:
The problem statement informs that:
Given:
- A box contains 100 balls from 1 to 100.
- 3 balls are selected randomly from the box.
Find Out:
- The probability that the sum of three numbers on the balls selected from the box will be odd.
Since we have 100 balls, there are 50 odd and 50 even balls.
As three balls are picked randomly from the box, there may occur two cases.
Case 1: if all three numbers are odd then, we get,
50* 50* 50 = 125* 1000
Or, Case 2: If two numbers are even, and one number is odd (EOE, EEO or OEE => 3 cases), then we get,
3* (50* 50* 50) = 3 * 125 *1000 = 375* 1000
=>Favourable cases = 125* 1000 + 375* 1000
= 1000* 500
Total cases= 100* 100* 100 = 1000*1000
Therefore, Probability = (1000* 500)/ (1000*1000) = ½
Thus, the probability that the sum of three numbers on the balls selected from the box will be odd is equal to ½.
Correct Answer: (C)
Approach Solution 3:
The problem statement informs that:
Given:
- A box contains 100 balls from 1 to 100.
- 3 balls are selected randomly from the box.
Find Out:
- The probability that the sum of three numbers on the balls selected from the box will be odd.
To find the sum of three numbers on the balls chosen from the box to be odd, one needs to choose three odd-numbered balls.
That is, (Odd + Odd + Odd = Odd)
Or, to find the sum of three numbers on the balls chosen from the box to be odd, one must choose two even-numbered balls and one odd-numbered ball.
That is (Even + Even + Odd = Odd)
P(OOO) = (1/2)^3 = 1/8
P (EEO) = 3* (1/2)^2*1/2 = 3/8 ( 3 is multiplied because the scenario presents two even-numbered balls and one odd-numbered ball that can happen in 3 distinct ways: EEO, OEE or EOE)
Therefore, we get, P= ⅛ + ⅜ = ½
Thus, the probability that the sum of three numbers on the balls selected from the box will be odd is equal to ½.
Correct Answer: (C)
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