Math, asked by shSaddam485, 1 year ago

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 Gagan555
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
Answered by DevendraLal
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)= \frac{number of red balls}{total number of balls}

                                                 P(R)  =\frac{10}{22}\\\\=\frac{5}{11}

probability of getting green ball P(G) = \frac{number of green balls}{total number of balls}\\\\=\frac{7}{22}

hence the probability of not getting a blue ball can also be written as the probability of RED+ GREEN

hence P(E)=  P(R)+P(G)

                 = \frac{5}{11}+ \frac{7}{22} \\\\

                 = \frac{10+7}{22} \\\\

                  =\frac{17}{22}

HENCE THE PROBABILITY OF A BALL DRAWN THAT IS NOT BLUE IS \frac{17}{22}.

               

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