A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being not a blue ball.
Answers
Answered by
46
P(E1)= Total no. of favorable conditions
-------------------------------------------------
Total no. of conditions
= No. of Red balls + No. of Green balls
--------------------------------------------------------
Total no. of balls
= 10 + 7
--------------
10 + 5 + 7
= 17
---
22
-------------------------------------------------
Total no. of conditions
= No. of Red balls + No. of Green balls
--------------------------------------------------------
Total no. of balls
= 10 + 7
--------------
10 + 5 + 7
= 17
---
22
Answered by
2
GIVEN,
number of balls red=10, blue=5, green=7
TO FIND,
probability of ball drawn not blue.
SOLUTION,
total number of balls = red balls+ blue balls+ green balls
= 10+5+7
= 22
probability of getting red ball P(R)=
P(R)
probability of getting green ball P(G)
hence the probability of not getting a blue ball can also be written as the probability of RED+ GREEN
hence P(E)= +
=
=
=
HENCE THE PROBABILITY OF A BALL DRAWN THAT IS NOT BLUE IS .
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