A Bag Contains 2 Red, 3 Green And 2 Blue Balls. Two Balls Are Drawn A

1.	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» 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 =7C2 = 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 =5C2= 10 Therefore, P(E) = n(E)/n(S) = 10/ 21. 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 =7C2 = 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 =5C2= 10 Therefore, P(E) = n(E)/n(S) = 10/ 21.

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»

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 =7C2 = 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 =5C2= 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

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 =7C2 = 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 =5C2= 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.