A bag contains \(x\) blue chips and \(y\) red chips. If the probability of selecting a red chip at random is \(\dfrac37,\) then \(\dfrac{x}{y} =\)
A. \(\dfrac7{11}\)
B. \(\dfrac34\)
C. \(\dfrac74\)
D. \(\dfrac43\)
E. \(\dfrac{11}7\)
[spoiler]OA=D[/spoiler]
Source: Magoosh
A bag contains \(x\) blue chips and \(y\) red chips. If the probability of selecting a red chip at random is
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We can quickly solve this question by testing values that satisfy the given informationVincen wrote: ↑Sat May 30, 2020 1:54 pmA bag contains \(x\) blue chips and \(y\) red chips. If the probability of selecting a red chip at random is \(\dfrac37,\) then \(\dfrac{x}{y} =\)
A. \(\dfrac7{11}\)
B. \(\dfrac34\)
C. \(\dfrac74\)
D. \(\dfrac43\)
E. \(\dfrac{11}7\)
[spoiler]OA=D[/spoiler]
Source: Magoosh
GIVEN: The probability of selecting a red chip at random is 3/7
So, it COULD be the case that there are 3 red chips, and a TOTAL of 7 chips
If there are 7 chips and 3 are red, then the other 4 chips must be blue
In other words, y = 3 and x = 4
This means x/y = 4/3
Answer: D
Cheers,
Brent