Question: There were r red balls and y yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out two balls at random, and puts the first ball back before pulling the second ball?
- \(\frac{(y+4)}{(y+r+7)} ∗ \frac{(r+3)}{(y+r+7)}\)
- \(\frac{(y+3)}{(y+r+7)} + \frac{(r+4)}{(y+r+6)}\)
- \(\frac{(y+3)}{(y+r+7)} ∗ \frac{(r+4)}{(y+r+6)}\)
- \(\frac{(y+3)}{(y+r+7)} ∗ \frac{(r+3)}{(y+r+7)}\)
- \(\frac{(y+3)}{(y+r+7)} + \frac{(r+3)}{(y+r+6)}\)
Solutions and Explanation
Approach Solution : 1
Amount of total red balls = r + 3,
Amount of total red balls = r + 3,
Therefore, the amount of total number of balls= r+3+y+4= r+y+7
The probability that a yellow ball will be chosen = (y+4)/(y+r+7)
He pulls the second ball after returning the first,
So the total number of balls again = r+y+7
The probability that a red ball will be chosen = (r+3)/(y+r+7)
The probability of them happening together = [(y+4)/(y+r+7)] * [(r+3)/(y+r+7)].
Correct Answer: (A)
Approach Solution : 2
Assume that the bag is empty prior to the addition of the 3 red balls and 4 yellow balls by setting r=0 and y=0.
P (yellow ball) = 4/7 (Four of the seven balls are yellow)
P(red ball) equals 3/7 ( Three of the seven balls are red
We multiply these probabilities to combine them, 4/7 * 3/7
The proper answer must produce something above the product when r=0 and y=0
Of all the options, only the equation in the first option can correspond to this,
[(y+4)/(y+r+7)] ∗ [(r+3)/(y+r+7)] = [(0+4)/(0+0+7)] ∗ [(0+3)/(0+0+7)] = (4/7) * (3/7)
Correct Answer:: (A)
“There were r red balls and y yellow balls in a bag. Three red balls” - is a subject covered in the GMAT quantitative reasoning section. A student needs to be knowledgeable in a wide range of qualitative skills in order to successfully complete GMAT Problem Solving questions. There are 31 questions in the GMAT Quant section overall. Calculative mathematical problems must be solved in the GMAT quant topics' problem-solving section using appropriate mathematical skills.
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