Question

You have a bag full of apples. There are 12 Jonathan (red), 6 Granny Smith (green), and 3 Golden Delicious (yellow) apples in the bag. The apples are all of similar size and shape. You reach into the bag (without looking) and pull out the number of apples listed in each scenario below.
a. Select 3 apples with replacement. Find the probability that all three are green.
b. Select 3 apples with replacement. Find the probability that none are red.
C. Select 4 apples without replacement. Find the probability that you get at LEAST 2 red apples.
d. Select 2 apples without replacement. Find the probability that the second apple is yellow given that the first apple is red.

Answer 1

Answer:

a) 0.032

b) 0.063

c) 0.313

d) 0.086

Step-by-step explanation:

12 red apples, 6 green apples and 3 yellow.

We have 21 apples!

a) the probability is

$\frac{\frac{6!}{3!(6-3)!}}{\frac{21!}{3!(21-3)!}}=\frac{6!3!18!}{3!3!21!}=0.032$

(all combnation to select tree green apples of 6 green apples derived with number of combinations to select 3 apples from all 21 apples)

b) probability of none are red is probability of chose 3 apples of 9 (sum of green and yellow).

$\frac{{9\choose 3}}{21\choose 3}=0.063$

c) least 2 red apples mean it will be 2 red, or 3 red or 4 red, so it is a sum of these tree probabilities.

$\frac{{12 \choose 2}}{{21 \choose 2}}*\frac{{9\choose 2}}{{19\choose 2}}+\frac{{12 \choose 3}}{{21 \choose 3}}*\frac{{9\choose 1}}{{18\choose 1}}+\frac{{12 \choose 4}}{{21 \choose 4}}$

$=0.15+0.08+0.083=0.313$

* expl: we have product because to chose 2 red apples has a probability and to chose 2 not red apples has a probabilit $\frac{{9\choose 2}}{{19\choose 2}}$  (9 sum of not red apples and 19 is 21 - chosen 2 red apples) *

d) It isa product of probability that you chose 1 red apple of 21 apples, and to chose 1 yellow apple of 20 apples

$\frac{{12\choose 1}}{{21\choose 1}}*\frac{{3\choose 1}}{{20\choose 1}}=0.086$

Answer 2

Answer:

The total number of apples is 17

Step-by-step explanation:

(a)

Number of green apples = 4

P ( that all three apples are green)

= (limit 4 and 3) / (limit 17 and 3)

= 4! / 3!(4−3)!

= 17! / 3!(17−3)!

= 1 / 170

= 0.005

(b)

Number of red apples = 8

P(that all three apple are green)

=(limit 8 and 3) / (limit 17 and 3)

= 8! / 3!(8−3)!

= 17! / 3!(17−3)!

= 7 / 85

= 0.082

Now,

P(that no three apple is red)= 1−0.082 = 0.918

c)

P(of selecting 4 apple that contain at least 2 red apples)

= (limit 8 and 2) /( limit 17 and 2) X (2/15) + (limit 8 and 3) /(limit 17 / 3) X (1/14)

= (7/34) × (2/15) + (7/85) × (1/14)

= 0.027 + 0.058

= 0.085

d)

P(that the second apple selected is yellow given that the first apple is red)

= (limit 5 and 1) / ( limit 16 and 1)

= 5! / 1!(5−1)!

= 16! / 1!(16−1)!

= 5 /16

= 0.3125