Answer:

You are supposed to draw 2 balls at random out of 10 balls wherein atleast one has to be even.

Also note: here there are 5 even balls and 5 odd balls. And they have to drawn with replacement policy..i.e you put the ball again into bag after removing each time.

So,

Atleast 1 even = either you pick one even OR you pick both even →#

So probabilty for having a single even number= either you pick first number as even AND the second number as odd OR you pick first number as odd AND second number as even=

(5/10). (5/10) + (5/10). (5/10) = 25/100

Now the probabilty for picking up both of the 2 balls as even= (5/10).(5/10)-(25/100)

Now as per ➡ #:

The total probabilty is =

(25/100).2+(25/100)

=3.(25/100)

=75/100

=0.75

= 75 % chance

1. A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0.

2. 140 times

3.5 drawn balls can have even sum if

• 1 even and 4 odd numbered balls
• 3 even and 2 odd numbered balls
• 5 even numbered balls

Have a good day!