Question

the probability that the coin
is not a 1 coin
is a <3 coin
3. A box has 4 red balls and 12 black balls Find the
4
probability that the selected ballis
a red ball
a black ball, chosen at random from the box
Also prove that the sum of these two pro​

Answer 1

Step-by-step explanation:

Let n

1

and n

2

be the number of red and black balls, respectively, in box I. Let n

3

and n

4

be the number of red and black balls, respectively, in box II.

Let E

3

be the event that transferred ball is red and E

4

be the event that transferred ball is black.

R be the event of selecting a Red ball.

P(R)=P(E

3

).P(R/E

3

)+P(E

4

).P(R/E

4

)

⇒P(R)=

3

1

=

n

1

+n

2

n

1

.(

n

1

+n

2

−1

n

1

−1

)+

n

1

+n

2

n

2

.(

n

1

+n

2

−1

n

1

)=

n

1

+n

2

n

1

only options C and D satisfies the above relation.

How satisfied