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