GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are $12$ red candies, $12$ blue, $7$ green, $13$ brown, $3$ orange, and $10$ 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 $=57$

Probability that first ball is orange, $P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{3}{57}=\frac{1}{19}$

Probability that second ball is orange, $P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}$

$=\frac{2}{56}=\frac{1}{28}$

Probability that third ball is green, $P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}$

$=\frac{7}{55}$

Now, probability that first two balls are orange and third is green is

$=P(A)\times P(B)\times P(C)$

$=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}$

$=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}$

$=\frac{1}{4180}$

Hence, probability that first two balls are orange and third is green is $\frac{1}{4180}$

3 0

4 years ago

(HELP 10 POINTS) Find the difference. Write your answer in standard form.

makkiz [27]

9514 1404 393

Answer:

-5x² -4x +8

Step-by-step explanation:

Eliminate parentheses and collect terms.

(-2x² +3) -(3x² +4x -5)

= -2x² +3 -3x² -4x +5

= (-2-3)x² -4x +(3 +5)

= -5x² -4x +8

4 0

3 years ago

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