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You reach into a bag that contains the following distribution of colored marbles

ID: 3301095 • Letter: Y

Question

You reach into a bag that contains the following distribution of colored marbles: 5 green, 8 yellow, 3 purple. You will grab 2 marbles (one at a time) and leave them on the table in front of you.

a) (1pt) Find the probability that you first select a green and second a yellow marble.

b) (1.5pts) Find the probability that you select no green marbles in your two picks.

c) (2pts) Find the probability that you select exactly 1 purple marble and one non-purple marble.

(2pts each) Mallory goes to school 4 days a week. She can either arrive on time or be late. Her tardiness on any given day is independent of all other days. If the probability that Mallory is late on any given day is 0.07…

a) …find the probability that she is on time for the first two days and tardy for the next two days.

b) …find the probability that she is late on at least one of the four days of the week.

Explanation / Answer

Question 1:

Total number of marbles in bag = 5 + 8 + 3 = 16

a) Probability that we first select a green marble and then a yellow marble is computed as:

= Probability of drawing a green marble * Probability of drawing a yellow marble given that a yellow marble is already drawn

= (5/16)*(8/15) = 1/6

Therefore (1/6) = 0.1667 is the required probability here

b) Now the probability that you select no green marbles in your two picks is computed as:

= [ (8+3)/ 16 ] * [ (11-1)/ 15) ]

= 0.4583

Therefore 0.4583 is the required probability here

c) Now the probability that you select exactly 1 purple marble and one non-purple marble is computed as:

= Probability that first is purple and second is non purple + Probability that first is non purple and second is purple

= [ (8 +5)/ 16] * [ 3/15] + [ 3/ 16] * [ (8+5) /15]

= 0.325

Therefore 0.325 is the required probability here.

Question 2:

a) Probability that she is on time for the first two days and tardy for the next two days

= ( 1- 0.07)2*0.072

= 0.00423801

Therefore 0.0042 is the required probability here.

b)Now probability that she is late on at least one of the four days of the week is computed as:

= 1 - Probability that she wont be late in any of the 4 days

= 1 - (1 - 0.07)4

= 0.25194799

Therefore 0.2519 is the required probability here.

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