HELP HELP HELP MATH GENIUSES, GODS AND GODDESSES
Suppose you are given 10 green balls labeled from 1-10, 5 red balls also labelled from 1-5 and 7 yellow balls labelled from 1-7. All of them are put in a bag and and you are asked to pick a ball once.
1. What is the probability that you will get a green ball labelled with a number less than 5 or an even number?
PLEASE NO NONSENSE :D
Share
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