Question #113218
A ball is drawn at random from a box containing 10 red balls, 30 white balls, 20 blue balls and 15 black balls.Obtain the probability that the ball drawn is:
(1)black or red
(2) not red or blue
(3)not blue
(4) white
(5) red, white,blue
1
Expert's answer
2020-05-01T18:14:29-0400

The total number of ball in the box , n(s)=10+30+20+15=75n(s)=10+30+20+15=75 .

Let, red ball denoted by RR

White ball denoted by WW

Blue ball denoted by BB

Black ball denoted by BlBl .


1) P(R orBl)=P(B)+P(Bl)P(R \ or Bl)=P(B)+P(Bl)


=1075+1575=2575=13=\frac{10}{75}+\frac{15}{75}=\frac{25}{75}=\frac{1}{3}


2) Let not red is denoted by R\sim R .

P(R orB)=P(R B)=P(R)+P(B)P(R B)P(\sim R \ or B)=P(\sim R \ \cup B)=P(\sim R)+P(B)-P(\sim R\ \cap B)

=1P(R)+P(B)P(B)=1-P(R)+P(B)-P(B) [since B (R)B\sub\ ( \sim R) ]

=11075=6575=1315=1-\frac{10}{75}=\frac{65}{75}=\frac{13}{15} .


3) P(not B)=1P(B)P(not \ B)=1-P(B)

=12075=5575=1115=1-\frac{20}{75}=\frac{55}{75}=\frac{11}{15} .


4) P(W)=3075=25P(W)=\frac{30}{75}=\frac{2}{5}


5) P(R W  B)=P(R)+P(W)+P(B)P( R \ \cup W\ \cup \ B)=P(R)+P(W)+P(B)

(As R,W,B are disjoint)

=1075+3075+2075=6075=45=\frac{10}{75}+\frac{30}{75}+\frac{20}{75}=\frac{60}{75}=\frac{4}{5} .


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