Question

There is a box containing 10 Ping Pong balls of three different colours red green and blue there are two red balls and 3 Green balls what is the probability of getting a blue balls in the first attempt

Answer 1

warning This is other question solution

Tom has 15 ping-pong balls each uniquely numbered from 1 to 15 also has a red box, a blue box and a green box.

(a) How many ways Tom place the 15 distinct balls into the three boxes so that no box is empty?

(b) Suppose now that Tom has places 5 ping-pong balls in each box. How many ways can he each 5 balls from the three boxes so that he chooses at least one from each box?

ANSWER

(a) No. of ways=

15

C

3

×3!×3

12

No. of ways to select three balls→

15

C

3

Placing the three balls such that one in each box →3!

Placing in other boxes→3

12

(b) Ways of collecting balls from the boxes:

1

(3R,1B,1G)

,

2

(1R,3B,1G)

,

3

(1R,1B,3G)

,

4

(2R,2B,1G)

,

5

(2R,1B,2G)

,

6

(1R,2B,2G)

No. of ways=

5

C

3

×

5

C

1

×

5

C

1

×3(1,2,3)+

5

C

2

×

5

C

2

×

5

C

1

(4,5,6)

=750+1500=2250.

Answer 2

The correct answer is $\frac{1}{2}$.

Given: Total balls in box = 10.

Number of Red balls = 2.

Number of Green balls = 3.

To Find: Probability of getting blue balls in the first attempt.

Solution:

Number of blue balls = Total balls - (red balls + green balls)

Blue balls = 10 - (2+3)

Blue balls = 5

Probability of getting a blue balls in the first attempt P(E) = $\frac{Favorable outcomes}{Total outcomes}$

P(E) = $\frac{Blue balls}{Total balls}$

P(E) = $\frac{5}{10}$

P(E) = $\frac{1}{2}$

Hence, the probability of getting a blue balls in the first attempt is $\frac{1}{2}$.

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