Exercise- CONDITIONAL PROBABILITY - #2
Solution:
A Coin can land as Head (H) or Tail (T); two choices. With three tosses, we can have belwo combinations:
2 x 2 x 2 = 8 combinations as below
1. H H H
2. H H T
3. H T H
4. H T T
5. T H H
6. T H T
7. T T H
8. T T T
We can now solve the below as:
What is the probability that exactly two heads occur, given that
(a) the first outcome was a head?
==> We have **two** combinations with exactly two heads with first outcome as head - H H T and, H T H.
==> P(a) = 2/8 = 0.25
(b) the first outcome was a tail?
==> We have **one** combination with exactly two heads with first outcome as tail - T H H.
==> P(b) = 1/8 = 0.125
(c) the first two outcomes were heads?
==> We have **one** combination with exactly two heads with first two outcomes as heads - H H T.
==> P(c) = 1/8 = 0.125
(d) the first two outcomes were tails?
==> We will have **zero** combinations with exactly two heads with first two outcomes as tails.
==> P(d) = 0/8 = 0.0
(e) the first outcome was a head and the third outcome was a head?
==> We have **one** combinations with exactly two heads with first and the third outcome as a head - H T H.
==> P(e) = 1/8 = 0.125