Respuesta :
Answer:
[tex]\frac{1}{4180}[/tex]
Step-by-step explanation:
GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are [tex]12[/tex] red candies, [tex]12[/tex] blue, [tex]7[/tex] green, [tex]13[/tex] brown, [tex]3[/tex] orange, and [tex]10[/tex] yellow. Three candies are pulled from the bag in succession, without replacement.
TO FIND: What is the probability that the first two candies drawn are orange and the third is green.
SOLUTION:
Total candies in the bag [tex]=57[/tex]
Probability that first ball is orange, [tex]P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}[/tex]
[tex]=\frac{3}{57}=\frac{1}{19}[/tex]
Probability that second ball is orange, [tex]P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}[/tex]
[tex]=\frac{2}{56}=\frac{1}{28}[/tex]
Probability that third ball is green, [tex]P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}[/tex]
[tex]=\frac{7}{55}[/tex]
Now, probability that first two balls are orange and third is green is
[tex]=P(A)\times P(B)\times P(C)[/tex]
[tex]=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}[/tex]
[tex]=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}[/tex]
[tex]=\frac{1}{4180}[/tex]
Hence, probability that first two balls are orange and third is green is [tex]\frac{1}{4180}[/tex]
