a Split each side of the equality into its individual parts using the formula for dependent events.
B
b Use the expressions we found in Part A to rewrite the inequality.
A
a 40 pink mints
B
b 41 pink mints
Practice makes perfect
a We are given that a bag holds 20 yellow mints and 80 mints that are green or pink. We have 100 mints in total. We are asked to find the number of pink mints if after eating a mint at random and then choosing another we have the following equality.
P(yellow then pink) = P(green then yellow)
Since we do not replace the mint, we have dependent events. Additionally, we can see that both sides of the given equality also represent dependent events. Recall that if A and B are dependent events, P(AthenB) is given as follows.
P(AthenB)= P(A)* P(BafterA)Using this information, we can split each side of the given equality. To do so, let's first let x be the number of pink mints and 80-x be the number of green ones. Now, let's first split the left side of the given equality. Keep in mind that after removing a mint we have 99 remaining mints.
P(yellow then pink) = P(yellow)* P(pink after yellow)
⇓
P(yellow then pink)=20/100* x/99
In the same way, we can split the right side of the equality.
P(green then yellow) = P(green)* P(yellow after green)
⇓
P(green then yellow)=80-x/100* 20/99
Now, using the above information we can rewrite the given equality.
20/100* x/99=80-x/100* 20/99
⇓
20 x/9900=20( 80-x)/9900
By solving this equation we will find the number of pink mints. Let's do it.
P(yellow then pink) > P(green then yellow)We are asked to find the least number of pink mints that keeps the given inequality true. Using the expressions we found in Part A, let's first rewrite the inequality.
20x/9900>20(80-x)/9900
Now, let's solve the inequality.
This means that the number of pinks must be greater than 40. The first whole number greater than 40 is 41. Therefore, the least number of pink mints is 41.
