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6 balls are drawn with replacement from a bag containing 7 red balls and 3 black balls. (round your answers to five decimal places.)

(a) Find the probability that 2 of the balls will be red. (b) Find the probability that all 6 balls will be black. (c) Find the probability that at least 4 of the balls will be black.

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Explanation:

"with replacement". that means any pulled ball is returned into the bag, and we have the full 7+3 = 10 balls to pull from every time.

remember, a probability is always

desired cases / totally possible cases

(a) 2 back will be red. I understand this, that exactly 2 balls will be red (not fewer, not more).

to get exactly 2 red balls we get 2 red balls and 4 black balls.

the probability for such a single event is

7/10 × 7/10 × 3/10 × 3/10 × 3/10 × 3/10 = 49×81/1000000 =

= 0.003969

because the probably to get a red ball (one from 7 out of 10) is 7/10.

the probability to get a black ball is 3/10.

how many different such single events are possible ?

as many as there are combinations to pick 2 out of 6 :

C(6, 2) = 6! / (2! × (6-2)!) = 6! / (2 × 4!) =

= 6×5 / 2 = 3×5 = 15

so, the probability to get exactly 2 red balls is

15 × 0.003969 = 0.059535 ≈ 0.05954

(b)

the probability of all 6 balls being black is

3/10 × 3/10 × 3/10 × 3/10 × 3/10 × 3/10 =

= 3⁶/1000000 = 0.000729 ≈ 0.00073

(c)

"at least 4 black balls" means exactly 4, or exactly 5, or exactly 6 balls are black.

that means we need to calculate all 3 probabilities and add them (non-overlapping "or"-cases means addition of probabilities, "and" cases mean multiplication as above).

to get exactly 4 black balls (= 4 black balls and 2 red balls) is the same as (a) : 0.059535

to get exactly 5 black balls (= 5 black balls and 1 red ball) in one specific constellation is

3⁵/10⁵ × 7/10 = 243×7/1000000 = 0.001701

and we have C(6, 5) possible constellations

C(6, 5) = 6! / (5! × (6-5)! = 6! / 5! = 6

the total probability to get exactly 5 black balls is

6 × 0.001701 = 0.010206

to get exactly 6 black balls is (b) : 0.000729

the probability to get at least 4 black balls is

0.059535 + 0.010206 + 0.000729 =

= 0.07047

User Milligran
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