A bag contains only red, blue, and green chips. One chip is drawn at

Question

A bag contains only red, blue, and green chips. One chip is drawn at random from the bag and put back, then again one chip is drawn at random from the bag and put back. The probability that both chips are red or that both chips are blue is 5/16. The probability that the first chip is red and the second chip is not red is 1/4. If one chip is drawn from the bag at random, what is the probability to draw a blue chip?

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

Are You Up For the Challenge: 700 Level Questions

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

Are You Up For the Challenge: 700 Level Questions

nick1816 · Accepted Answer

Number of Red chips =R

Number of Blue chips =B

Number of Total chips =T

\frac{R^2}{T^2} + \frac{B^2}{T^2} =\frac{ 5}{16} ......(1)

Also,
 \frac{R}{T} (1-\frac{R}{T}) = \frac{1}{4}

\frac{R}{T} = \frac{1}{2} .......(2)

From (1) and (2)

\frac{B^2}{T^2} = \frac{1}{16}

\frac{B}{T} = \frac{1}{4}

sambitspm · Answer

P(r) = Probability of red ball in one pick.
P(r)*P(r) + P(b)*P(b) = 5/16

Now P(r)  = 1/4

=> P(r)= 1/2

So keeping first equation
1/4 + P(b)^2 = 5/16
P(b) = 1/4

C

ShreyasJavahar · Answer

Could you please explain how you arrived at equation 2?

nick1816 · Answer

ShreyasJavahar

probability to draw a Red chip = Number of red chips/ total number of chips = R/T

probability to draw a chip that is not Red = 1- R/T

Since we are replacing back the first chip, the second draw is independent of the first draw.

Hence P(A and B) = P(A) * P(B)

The probability that the first chip is red and the second chip is not red = (R/T) * (1- R/T)

Getsome2 · Answer

nick1816

Thank you. The formula for Part 2 (and I) makes sense. I must just be dense rn but how did you solve for R/T as 1/2 then? Thank you

nick1816 · Answer

\frac{R}{T} (1-\frac{R}{T}) = \frac{1}{4}

Assume R/T =x

x(1-x) = \frac{1}{4}

x-x^2 = \frac{1}{4}

x^2-x +\frac{1}{4} = 0

x^2 - (2*x*\frac{1}{2}) +(\frac{1}{2})^2 = 0 ...... x= 2*\frac{1}{2}*x ]

(x-\frac{1}{2})^2 = 0 ........ (a-b)^2 = a^2 - 2ab +b^2 ]

x= \frac{1}{2}

If you still got any doubt, you can ask.

Getsome2 · Answer

Ah I understand now. Thank you for explaining and your response.

Lipun · Answer

(R^2 + B^2)/T^2 = 5/16

Consider T = 4, => R = 2 , B = 1 or R = 1, B = 2 (since R^2 + B^2 = 5)

If R = 2, B = 1, Not R = 4-2 = 2
=> P(R)*P(Not R) = 2/4 * 2/4 = 1/4 => P(B) = 1/4

If R = 1, B = 2, Not R = 4-1 = 3
=> P(R)*P(Not R) = 1/4 * 3/4 = 3/16 (contradicts given statement)

=> P(B) = 1/4

Ans: C

Louis14 · Answer

Can you tell how you got P(r) = 1/2 from P(r)  = 1/4?

Alvaro20172027 · Answer

Being r=red, b=blue, y=yellow and r+b+y=x
1. P(pick 2 same colour)= r/x*r/x + b/x*b/x= 5/16
 (r^2+b^2)/x^2=5/16
 we can infere than x=4, then r and b can be 1 or 2.

2. P(first red then not red)= r/x+(b+y)/x^4=1/4, 
 as we previously found x value, then the sum in the numerator equals to 4
 r(b+y)=4

Then, if a and b are 1 or 2, but the cant be both the same, a+b=3. So y=1
 So r(b+1)=4
 The only possible value of r and b, with the a+b=3 restriction is r=2 and b=1

P(pick b)= 1/4

Kinshook · Answer

Given: A bag contains only red, blue, and green chips. One chip is drawn at random from the bag and put back, then again one chip is drawn at random from the bag and put back. The probability that both chips are red or that both chips are blue is 5/16. The probability that the first chip is red and the second chip is not red is 1/4.

Asked: If one chip is drawn from the bag at random, what is the probability to draw a blue chip?

Let red, blue and green chips be R, B & G respectively
T = R + B + G

One chip is drawn at random from the bag and put back, then again one chip is drawn at random from the bag and put back. The probability that both chips are red or that both chips are blue is 5/16. 
R/T * R/T + B/T * B/T = 5/16
Rˆ2/Tˆ2 + Bˆ2/Tˆ2 = 5/16

The probability that the first chip is red and the second chip is not red is 1/4. 
R/T * (1 - R/T) = 1/4
R/T = 1/2

Bˆ2/Tˆ2 = 5/16 - 1/4 = 1/16
B/T = 1/4

IMO C

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