For a certain candy,20% of the pieces are yellow,15% are red,55% are blue,20% ar

Question

For a certain candy,20% of the pieces are yellow,15% are red,55% 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?

a) The probability that it is brown is nothing.

(Round to three decimal places as needed.)

The probability that it is yellow or blue is nothing.

(Round to three decimal places as needed.)

The probability that it is not green is nothing.

(Round to three decimal places as needed.)

The probability that it is striped is nothing.

(Round to three decimal places as needed.)

b) The probability of picking three brown candies is nothing.

(Round to three decimal places as needed.)

The probability of the third one being the first red one is nothing.

(Round to three decimal places as needed.)

The probability that none are yellow is nothing.

(Round to three decimal places as needed.)

The probability of at least one green candy is nothing.

(Round to three decimal places as needed.)

Explanation / Answer

Solution:

For a certain candy, 20% of the pieces are yellow, 15% are red, 5% 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?
P(brown) = 40/100 = 0.4
yellow or blue? = 20/100 + 5/100 = 25/100 = 0.25
not green? = 1 - P(green) = 1 - 20/100 = 0.8
striped?
b) Assume you have an infinite supply of these candy pieces from which to draw. If you pick up three pieces in a row, what is the probability that they are
brown? = P( all three are brown) = 0.4^3 = 0.064
third one being the first red one? = P( third is the red) = 0.8*0.8*0.2 = 0.128
none are yellow? = 1 - P(Yellow ) = 1 - (0.20)^3 = 0.992
at least one green? = 1-P[no green] = 1-(0.80)^3 = 0.488

P(Yellow) = 0.20
P(red)=0.15
P(blue)=0.05
P(green)=0.20
a)P(brown)=1-(0.20+0.15+0.05+0.20)=0.40
b)P(yellow or blue)=P(Yellow)+P(blue)= 0.25
c)P'(green)=1-P(green)=1-0.20=0.80
d)P(striped)=0
e)P(3 brown)=0.40^3= 0.064
f)P( third one being the first red one)=P'(red)*P'(red)*P(red) = 0.15*0.15*0.85= 0.019
g)P(none yellow)=P'(yellow)^3=(1-0.20)^3= 0.512
h)P(aleast 1 green)=1-P(no green)=1-(0.20)^3= 0.992

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