MA22S6 Homework 1 Solutions
ˆ Question 1
The obvious sample space is that enumerating every possible outcome, namely
{HHH, HHT, HT H, T HH, T T H, T HT, HT T, T T T }. Other sample space listings are
{0 tails, 1 tail, 2 tails, 3 tails} and {2 heads or more, less than 2 heads}. Now we want to
calculate the probability of i tails, for i = 0, 1, 2, 3. The event of i tails is written as
Ai , given by P(Ai ) = (3i )(0.5)i (0.5)3−i . Plugging in the various values for i gives P(A0 ) =
P(A3 ) = 0.125 and P(A1 ) = P(A2 ) = 0.375. We use Bayes’ Theorem to find the probability
of the second toss yielding a tail given that two of the three tosses yielded heads:
P(2 Heads∣2nd T )P(2nd T )
,
P(2 Heads)
P(1st H and 3rd H)P(2nd T )
=
,
P(A2 )
(0.5 × 0.5)(0.5) 1
= .
=
0.375
3
P(2nd T∣2 Heads) =
Finally P(E) = P(A0 ) + P(A1 ) + P(A2 ) = 87 .
ˆ Question 2
In the first part we denote by C the event of a plant being sprayed by a chemical, by
C̄ the event that is not, by D the event that a plant dies within the month, and by D̄
the event that it does not. We write this information as: P(C) = 0.01, P(C̄) = 0.99,
P(D∣C) = 0.9 → P(D̄∣C) = 0.1 and P(D∣C̄) = 0.1 → P(D̄∣C̄) = 0.1. Using Bayes we find
P(D∣C)P(C)
P(D)
P(D∣C)P(C)
=
P(D∣C)P(C) + P(D∣C̄)P(C̄)
0.9 × 0.01
=
= 0.0833
0.9 × 0.01 + 0.1 × 0.99
P(C∣D) =
In the next part we denote by O the event of an area containing oil, by Ō the event that is
not, by + the event that the test says the are contains oil, and by − the event that it does
not. We can now write the following given the information we’re provided: P(O) = 0.02,
1
¯ = 0.98, P(+∣Ō) = 0.1 → P(−∣Ō) = 0.9 and P(−∣0) = 0.05 → P(+∣O) = 0.95. By Bayes
P(O)
P(−∣O)P(O)
P(−)
P(−∣O)P(O)
=
P(−∣O)P(O) + P(−∣Ō)P(Ō)
0.05 × 0.02
=
= 0.00113.
0.05 × 0.02 + 0.9 × 0.98
P(O∣−) =
(1)
ˆ Question 3 Remark: a simultaneous draw is technically the same as a consecutive draw
without replacement.
– (a)
3
8
× 27 =
– (b) ( 38 )3 =
– (c)
3
8
3
28 .
1
56 .
× 27 × 16 =
27
512 .
– (d) P(2 out of 3R) = P(RRB)+P(RBR)+P(BRR)=( 38 × 72 × 65 )+( 38 × 57 × 26 )+( 58 × 37 × 26 ) =
15
56 .
2