Respuesta :
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
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
[tex]\frac{\frac{6!}{3!(6-3)!}}{\frac{21!}{3!(21-3)!}}=\frac{6!3!18!}{3!3!21!}=0.032[/tex]
(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).
[tex]\frac{{9\choose 3}}{21\choose 3}=0.063[/tex]
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.
[tex]\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}}[/tex]
[tex]=0.15+0.08+0.083=0.313[/tex]
* expl: we have product because to chose 2 red apples has a probability and to chose 2 not red apples has a probabilit [tex]\frac{{9\choose 2}}{{19\choose 2}}[/tex] (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
[tex]\frac{{12\choose 1}}{{21\choose 1}}*\frac{{3\choose 1}}{{20\choose 1}}=0.086[/tex]
