Question

For a certain​ candy, 15​% of the pieces are​ yellow, 10​% are​ red, 15​% are​ blue, 20​% are​ green, and the rest are brown. ​a) If you pick a piece at​ random, what is the probability that it

is​ brown?

it is yellow or​ blue?

it is not​ green?

it is​ striped?

​b) Assume you have an infinite supply of these candy pieces from which to draw. If you pick three pieces in a​ row, what is the probability that they

are all​ brown?

the third one is the first one that is​ red?

none are​ yellow?

at least one is​ green?

Answer 1

GIVEN THAT

Probability of candies as below

yellow=15%=0.15

red=15%=0.15

blue=10%=0.10

green =15%=0.15

brown=1-(0.15+0.15+0.10+0.15)=1-0.55

=0.45

(a) If you pick a piece at random , there

The probability that it is brown

P(brown)=1-(0.15+0.15+0.10+0.15)=0.45

The probability that it is yellow as below

P(yellow blue)=P(yellow)+P(blue)

=0.15+0.10

=0.25

The probability that it is net green .Let X the required probability

P(X)=1-P(green)

=1-0.15=0.85

The probability that it is striped since there are no striped candies so the probability will is zero

(b) If you pick there pices is a row probability that they are all brown .

Probability of brown candidate it 0.45

Let A be the required probability

P(A)=P(brown)*P(brown)*P(brown)

=0.45*0.45*0.45

=0.0911

The probability of the third one is the first one that is red

since first two candidate are net red , the required probability will be

P(net red)=1-015=0.85

Let X be the required probability

P(X)=0.85*0.85*0.15

=0.1083

The probability that none are yellow. probability of yellow candies is 0.15

P(net yellow )=1-0.15=0.85

Let the required probability be Y

P(Y)=0.85*085*085

=0.6141

The probability that at least on is green probability of net green candies is 0.85

Let Z be the required probability

P(Z)=1-0.85*0.85*0.85= =1-0.6141

=0.3858