A candy jar contains the following candies:  7 cherry, 3 lemon, 4 lime, 6 orange.  TWO (2) candies are drawn WITHOUT REPLACEMENT.  Use this information to answer questions #4 - 6.4)  P(BOTH orange) =1 pointYour answer5)  P(cherry on the first draw AND lime on the second draw) =1 pointYour answer6)  P(lemon on the 2nd draw GIVEN THAT the first draw was orange) =1 point

Sure, let's solve these step by step.

4. P(BOTH orange)

First, calculate the total number of candies: Total candies = 7 (cherry) + 3 (lemon) + 4 (lime) + 6 (orange) = 20 candies

The probability of drawing an orange candy on the first draw: P(First orange) = Number of orange candies / Total number of candies = 6/20

After drawing one orange candy, there are now 19 candies left, and 5 of them are orange.

The probability of drawing an orange candy on the second draw: P(Second orange) = Number of remaining orange candies / Total number of remaining candies = 5/19

Therefore, the probability of both candies being orange is: P(BOTH orange) = P(First orange) * P(Second orange) = (6/20) * (5/19) = 30/380 = 3/38

Your answer: 3/38

5. P(cherry on the first draw AND lime on the second draw)

The probability of drawing a cherry candy on the first draw: P(First cherry) = Number of cherry candies / Total number of candies = 7/20

After drawing one cherry candy, there are now 19 candies left, and 4 of them are lime.

The probability of drawing a lime candy on the second draw: P(Second lime) = Number of remaining lime candies / Total number of remaining candies = 4/19

Therefore, the probability of drawing a cherry first and a lime second is: P(cherry on the first draw AND lime on the second draw) = P(First cherry) * P(Second lime) = (7/20) * (4/19) = 28/380 = 7/95

Your answer: 7/95

6. P(lemon on the 2nd draw GIVEN THAT the first draw was orange)

The probability of drawing an orange candy on the first draw is not needed here since we are given that the first draw was orange.

After drawing one orange candy, there are now 19 candies left, and 3 of them are lemon.

The probability of drawing a lemon candy on the second draw given that the first draw was orange: P(lemon on the 2nd draw GIVEN THAT the first draw was orange) = Number of remaining lemon candies / Total number of remaining candies = 3/19

Your answer: 3/19