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A jar contains 30 red marbles, 40 blue marbles, 50 green mar
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13 Oct 2020, 09:21
13 Oct 2020, 09:21
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Question Stats:
88% (01:34) correct
11% (02:18) wrong based on 51 sessions
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A jar contains 30 red marbles, 40 blue marbles, 50 green marbles, and no other marbles. After x blue marbles are added to the jar, the probability of selecting a blue marble on a random draw is equal to \(\frac{7}{9}\). What is the value of x?
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Show: :: OA
240
Kudos for the right answer and explanation
Re: A jar contains 30 red marbles, 40 blue marbles, 50 green mar
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14 Oct 2020, 07:09
14 Oct 2020, 07:09
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The probability of getting 40 blues marble is 40/120=1/3
The desire probability is 7/9
So we divide (7/9)/(1/3)= 7/9*3/1=7/3
So we multiply the numerator by 7 and denominator by 3
We got 40*7/120*3=240/360
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The desire probability is 7/9
So we divide (7/9)/(1/3)= 7/9*3/1=7/3
So we multiply the numerator by 7 and denominator by 3
We got 40*7/120*3=240/360
Posted from my mobile device
Re: A jar contains 30 red marbles, 40 blue marbles, 50 green mar
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05 Aug 2021, 11:40
05 Aug 2021, 11:40
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7/9=0.778
(40+x)/(90+x)=0.778
40+x=0.778*(90+x)
0.222x=90*0.778-40
0.222x=30
x=315.15
shouldn't be x=315?
(40+x)/(90+x)=0.778
40+x=0.778*(90+x)
0.222x=90*0.778-40
0.222x=30
x=315.15
shouldn't be x=315?
