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Direction: There are three boxes P, Q and R which consists of Red, Blue and Green colour balls. Total number of blue balls in all three boxes together is 9 and total number of balls in all the boxes together is 32.
In Box P: The probability of choosing a Blue ball is 1/4 and the number of red balls is 2 more than the number of blue balls.
In Box Q: Total number of balls is 10 and the ratio of probability of getting a Blue and a Red ball is 2 : 3. Also the sum of Blue and Green balls is 4 more than the number of red balls.
In Box R: Total number of balls is 10 and the probability of getting a red is 1/5 less than the probability of getting a blue.
One ball each is drawn from each box. Find the probability that all are of red colour.
1/40
1/24
3/50
4/15
1/30
Let’s break this problem into bite-sized pieces:
- Total blue balls in all boxes = 9
- Total balls across P, Q, R = 32
- Box P: Probability(Blue) = 1/4 ? Blue balls in P = Bp, so Bp/Np = 1/4 ? Np = 4Bp
- Red balls in P = Bp + 2
- Box Q: Total balls = 10. Let Blue in Q = Bq, Red in Q = Rq, Green in Q = Gq
- Probability ratio Blue:Red = 2:3 ? Bq/Rq = 2/3
- Bq + Gq = Rq + 4
- Box R: Total balls = 10. Probability(Red) is 1/5 less than Probability(Blue)
- Let Blue in R = Br, Red in R = Rr, Green in R = Gr
- Rr/10 = Br/10 – 1/5 ? Rr = Br – 2
Now, step-by-step math:
Box P:
- Let Blue balls in P = x
- Np = 4x
- Red balls = x + 2
- So Green balls = Np – (x + x + 2) = 4x – (x + x + 2) = 2x – 2
Box Q:
- Bq + Rq + Gq = 10
- Bq/Rq = 2/3 ? 3Bq = 2Rq ? Bq = (2/3)Rq
- Let Rq = y, then Bq = (2/3)y
- Bq + Gq = Rq + 4 ? (2/3)y + Gq = y + 4 ? Gq = y + 4 – (2/3)y = (1/3)y + 4
So, sum:
- (2/3)y + y + (1/3)y + 4 = 10
- (2/3 + 1 + 1/3)y + 4 = 10
- (2y) + 4 = 10 ? 2y = 6 ? y = 3
Thus, Rq = 3, Bq = 2, Gq = 5
Box R:
- Br + Rr + Gr = 10
- Rr = Br – 2
- So (Br – 2) + Br + Gr = 10 ? 2Br – 2 + Gr = 10 ? Gr = 12 – 2Br
We need to check all possibilities for blue balls in each box.
Total Blue balls = x (Box P) + 2 (Box Q) + Br (Box R) = 9
Let’s try:
x + 2 + Br = 9 ? x + Br = 7
From Box P, x must be integer and Np = 4x = 32 (Since total balls = 32), so try x values.
Try x = 3 ? Br = 4
x = 3
Np = 12
Red = 5
Green = 12 – 3 – 5 = 4
Blue = 2, Red = 3, Green = 5
Br = 4, Rr = 4 – 2 = 2, Gr = 10 – 4 – 2 = 4
Check total balls:
Box P: 12
Box Q: 10
Box R: 10
Sum = 12 + 10 + 10 = 32
Blue balls: 3 (P) + 2 (Q) + 4 (R) = 9
Everything fits. Now, find probability that all three are red:
- P (red in P) = Red balls in P / Total balls in P = 5/12
- P (red in Q) = 3/10
- P (red in R) = 2/10 = 1/5
So, overall probability = 5/12 × 3/10 × 1/5 = (5×3×1)/(12×10×5) = 15/600 = 1/40
Here’s what this means:
- Every part of the question lines up—each step flows into the next.
- The options are about calculating this "all red" scenario from the setup.
- Option 1 (1/40) is exactly the correct answer.
Correct Answer: Option 1 — 1/40
By: Parvesh Mehta ProfileResourcesReport error
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