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MAT 142 College Mathematics
Module PM
Worksheet 2
Terri Miller
revised February 23, 2011
1. Mendel found that snapdragons have no color dominance; a snapdragon with one red
gene and one white gene will have pink flowers. If a pure-red snapdragon is crossed
with a pure-white snapdragon, find the probability of the following.
Solution: Create a Punnett square, using w for white gene and R for red gene.
w
w
R
R
wR wR
wR wR
(a) a red offspring Solution: 0, since no offspring will be RR
(b) a white offspring Solution: 0, since no offspring will be ww
(c) a pink offspring Solution: 1, since all offspring will be wR
2. For diseases which are recessive, one gene makes the individual a carrier (but not ill)
and two genes makes the individual have the disease. Tay-Sachs disease is a recessive
disease. If carrier-detection tests show that one prospective parent is a carrier of
Tay-Sachs and the other has no Tay-Sachs gene, find the probability of each of the
following.
Solution: Create a Punnett square , using t for Tay-Sachs gene and N for non TaySachs gene
N
t
N
NN
tN
N
NN
tN
(a) The child would have the disease.
Solution: Since this requires tt and there are none, this probability is 0
(b) The child would be a carrier.
Solution: This is tN, there are two such offspring, so this probability is
2
1
=
4
2
(c) The child would be healthy (i.e., no gene for Tay-Sachs).
Solution: This is NN, so this is also
1
2
1
=
4
2
c 2011 ASU School of Mathematical & Statistical Sciences and Terri L. Miller
3. Use probability rules whenever possible (and then check by counting on the chart) to
determine the following. A pair of dice are rolled (one red and one green), what is the
probability that
(a) the sum of the pair is 9
Solution: Let E be the event that the pair sums to 9. This one must be done
directly, we see from the chart of rolls of two dice that there are 4 ways to sum
4
1
= .
to 9, so this probability is p(E) =
36
9
(b) the sum of the pair is greater than 9
Solution: Let F be the event that the sm is greater than 9. We still need
to do this one directly, we count the rolls that sum to 10, 11, and 12 to get
1
6
= .
p(F ) =
36
6
(c) the sum of the pair is not greater than 9
Solution: The event that the sum is greater than or equal to 9 is E ∪ F . Using
the formula p(E ∪ F ) = p(E) + p(F ), from (a) and (b) (note that no roll can be
4
6
both nine and greater than 9 at the same time), we have p(E ∪F ) =
+ −0 =
36 36
10
5
= .
36
18
(d) the sum of the pair is less than 9
Solution: The sum less that 9 is the complement of the sum greater than or
equal to 9, so using the formula p(E) + p(E) = 1 and the previous part, we have
5
13
1−
= .
18
18
(e) the sum of the pair is even
1
Solution: Since half of the sums are even, this probability is .
2
(f) the sum of the pair is greater than 9 and even
Solution: Let G be the event that the sum is greater than 9, and H be the event
1
4
that the sum is even. We will count this directly to get p(G ∩ H) =
= .
36
9
(g) the sum of the pair is greater than 9 or even
Solution: This is G∪H, so we use our formula, p(G∪H) = p(G)+p(H)−p(G∩H)
1 1 1
6
18
4
along with the previous result to get p(G ∪ H) = + − =
+
−
=
6 2 9
36 36 36
20
5
= .
36
9
(h) the difference is 4
Solution: We will need to count these directly. For the difference to be 4, one
die would need a 6 and the other a 2 (2 cases) or one would need a 5 and the
4
1
other a 1 (2 cases). This gives
= .
36
9
2
c 2011 ASU School of Mathematical & Statistical Sciences and Terri L. Miller