Question

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

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

=

7

C

2

=

(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.

=

5

C

2

=

(2×1)

(5×4)

=10

∴P(E)=

n(S)

n(E)

=

21

10

Answer 2

Step-by-step explanation:

\huge\underbracre\blue{Answer}

$\huge\underbrace\blue{Answer}$

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

n (S)= 7C²

n (S)=(7×6)/(2×1)

n (S) =21

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

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

n(E)= 5C²

n (E)=(5×4)/(2×1)

n (E) =10

: p (E) = n (E)/ n (S) =10/21

Hope its help