CRE 4 - Picking balls from a bag | Modern Math - Probability

Answer: Option B

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Explanation :

For both the balls to be of different color, we need to pick either 

1 blue & 1 red = 5 × 6 = 30 ways.
1 red & 1 black = 6 × 4 = 24 ways
1 blue & 1 black = 5 × 4 = 20 ways.

∴ Number of ways of picking 2 balls of different color = 30 + 24 + 20 = 74 ways.

Now, total number of ways of picking 2 balls out of 15 = ¹⁵C₂ = 105

P (both the balls are of different color) = 74/105

Hence, option (b).

Answer: Option B

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Explanation :

Let us first calculate P (1 ball is blue and 1 red) = P(1^st blue and 2^nd red) + P (1^st red and 2^nd blue).

P(1^st blue) = $\frac{5}{15}$

Now, there are 14 balls remaining.

P(2^nd red) = $\frac{6}{14}$

P(1^st blue and 2nd red) = $\frac{5}{15}\times \frac{6}{14}$

Similarly, P (1^st red and 2^nd blue) = $\frac{6}{15}\times \frac{5}{14}$

P (1 ball is blue and 1 red) = $\frac{5}{15}\times \frac{6}{14}$ + $\frac{6}{15}\times \frac{5}{14}$ = 60/210.

Similarly

P (1 ball is blue and 1 black) = $\frac{5}{15}\times \frac{4}{14}$ + $\frac{4}{15}\times \frac{5}{14}$ = 40/210

P (1 red is blue and 1 black) = $\frac{6}{15}\times \frac{4}{14}$ + $\frac{4}{15}\times \frac{6}{14}$ = 48/210

∴ P (both balls are of different color) = P (1 ball is blue and 1 red) + P (1 ball is blue and 1 black) + P (1 red is blue and 1 black) 

⇒ ∴ P (both balls are of different color) = $\frac{60}{210}+\frac{40}{210}+\frac{48}{210}$ = $\frac{148}{210}$ = $\frac{74}{105}$

Hence, option (b).

Answer: Option C

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Explanation :

Let us first calculate P (1 ball is blue and 1 red) = P (1^st blue and 2^nd red) + P (1^st red and 2^nd blue).

P (1^st blue) = 5/15

Now, this ball is again replaced, hence we again have a total of 15 balls to pick from.

P (2^nd red) = 6/15

P (1^st blue and 2^nd red) = 5$\frac{5}{15}\times \frac{6}{15}$

Similarly, P (1^st red and 2^nd blue) = $\frac{6}{15}\times \frac{5}{15}$

P (1 ball is blue and 1 red) = $\frac{5}{15}\times \frac{6}{15}$ + $\frac{6}{15}\times \frac{5}{15}$ = $\frac{60}{225}$.

Similarly

P (1 ball is blue and 1 black) = $\frac{5}{15}\times \frac{4}{15}$ + $\frac{4}{15}\times \frac{5}{15}$ = $\frac{40}{225}$

P (1 red is blue and 1 black) = $\frac{6}{15}\times \frac{4}{15}$ + $\frac{4}{15}\times \frac{6}{15}$ = $\frac{48}{225}$

∴ P (both balls are of different color) = P (1 ball is blue and 1 red) + P (1 ball is blue and 1 black) + P (1 red is blue and 1 black) 

⇒ ∴ P (both balls are of different color) = $\frac{60}{225}+\frac{40}{225}+\frac{48}{225}$ = $\frac{148}{225}$

Hence, option (c).

Answer: Option B

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Explanation :

For all balls to be of different color, we need to pick exactly 1 blue, 1 red and 1 black blall.

This can be done in 5 × 6 × 4 = 120 ways.

Now, total number of ways of picking 3 balls out of 15 = ¹⁵C₃ = 455

P (balls are of different color) = 120/455 = 24/91

Hence, option (b).

Answer: Option B

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Explanation :

Drawing 3 balls simultaneously is same as drawing 3 balls one by one without replacement.

∴ P (balls are of different color) = 24/91

Hence, option (b).

Answer: Option A

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Explanation :

Let’s first calculate P (1^st blue, 2^nd red & 3^rd black) = $\frac{5}{15}\times \frac{6}{15}\times \frac{4}{15}$ = $\frac{120}{3375}$

Now, the sequence of drawing three colors can be in 3! = 6 ways.

∴ We multiply the above probability with 6 to get the final answer.

∴ Required answer = $\frac{120}{3375}\times 6$ = $\frac{80}{375}$ = $\frac{16}{75}$

Hence, option (a).