Question

2. A bag contains 4 red balls, 7 white balls, 3 green balls and
6 black balls. If one ball is
drawn at random. Find the probability that it is
a) Black ball
b) Not white​

Answer 1

$\huge\pink{\boxed{\green{\mathfrak{\overbrace{\underbrace{\fcolorbox{red}{aqua}{\underline{\pink{✯Answer✯}}}}}}}}}$

a) 3/10

b) 13/20

Answer 2

Explanation:

Let E

1

and E

2

be the events of selecting first bag and second bag respectively.

∴P(E

1

)=P(E

2

)=

2

1

Let A be the event of getting a red ball.

⇒P(A∣E

1

)=P(drawing a red ball from first bag)=

8

4

=

2

1

⇒P(A∣E

2

)=P(drawing a red ball from second bag)=

8

2

=

4

1

The probability of drawing a ball from the first bag, given that it is red, is given by P(E

2

∣A).

By using Baye's theorem, we obtain

P(E

1

∣A)=

P(E

1

)⋅P(A∣E

1

)+P(E

2

)⋅P(A∣E

2

)

P(E

1

)⋅P(A∣E

1

)

=

2

1

⋅

2

1

+

2

1

⋅

4

1

2

1

⋅

2

1

=

4

1

+

8

1

4

1

=

8

3

4

=

3

2

=0.66