Question

Q2. A bag contains 2 red, 3 green and 2 blue balls. Two balls are
drawn at random. What is the probability that none of the balls
drawn is blue?

Answer 1

Answer:

Probability = 10/21. A bag contains 2 red, 3 green, and 2 blue balls.

Step-by-step explanation:

Total number of balls = (2 + 3 + 2) = 7

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7= (2×1)(7×6)

=21

Let E = Event of drawing 2 balls, none of which is blue.

∴n(E)= Number of ways of drawing 2 balls out of (2 + 3) balls.

-(2×1)

-(5×4) =10

∴P(E)= n(S)n(E)= 21/10

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Answer 2

2 RED: R1 R2

3 GREEN: G1 G2 G3

2 BLUE: B1 B2

2 balls are drawn randomly.

Starting with R1 which is drawn along with either of the remaining 6 balls. Means possible 6 ways for R1.… (1)

Then R2 is drawn with either of the remaining 5 balls. Means possible 5 ways for R2…….(2)

Similarly, we draw G1, G2, G3, B1, B2 . For each possible ways = 4, 3. 2, 1, 0 ………(3)

So, total possible ways = 6+5+4+3+2+1 = 21

So, this is how we calculate by permutation Combination too..

by 7C2 = combination of 7, objects taken 2

= 7! /{ (7–2) ! 2!} = 21 ●

Now excluding B1 B2 , we find the possible ways out.. Starting from R1, paired with R2, G1, G2, G3… ie 4 + 3 + 2+ 1+ 0 = 10

Or, 5C2 = 5! / {(5–2)! 2! } = 10 ●

Hence, probability( event, none of them blue) = (no of outcomes favourable to the event) /( total outcomes)

= 10/ 21 ●●● ANS