Frinz Danzelle Aedan A. Uy
BSIE-2
IE 2103 Friday 3:00 PM-6:00 PM
HOME ACTIVITY 1.1
1. An experiment has three outcomes, I, II, and III. If outcome I is twice as likely as
outcome II, and outcome II is three times as likely as outcome III, what are the
probability values of the three outcomes?
SOLUTION:
Seeing that there are three possible outcomes in this event, their probabilities should
add up to 1. As per the condition: P(I) = 2P(II) and P(II) = 3P(III).
Which gives us,
P(I)+P(II)+P(III) = 1.0
And following the conditions above, we have,
2(3P(III))+3P(III)+P(III) = 1.0
To find the value, we let P(III) = x
(2(3x)) + (3x) + x = 1
6x + 3x + x = 1
10x = 1
Which makes P(III) = 0.1
Therefore:
P(I) = 2(3(0.1)) = 0.6
P(II) = 3(0.1) = 0.3
And
P(III) = 0.1
2.
(a) P( B ) = 0.01 + 0.02+ 0.05 + 0.11 + 0.0 + 0.06 + 0.13 = 0.46
(b) P( B ∩ C ) = 0.02 + 0.05 + 0.11 = 0.18
(c) P( A ∪ C ) = 0.07 + 0.05 0.01 + 0.02 + 0.05 + 0.08 + 0.04 + 0.11 + 0.07 + 0.11 =
0.61
(d) P( A ∩ B ∩ C ) = 0.02 + 0.05 = 0.07
(e) P( A ∪ B ∪ C ) = 1 – 0.03 – 0.04 – 0.05 = 0.88
(f) P( A’ ∩ B ) = 0.08 + 0.06 + 0.11 + 0.13 = 0.38
(g) P( B’ ∪ C ) = 0.04 + 0.03 + 0.05 + 0.11 + 0.05 + 0.02 + 0.08 + 0.04 + 0.11 + 0.07 +
0.07 + 0.05 = 0.72
(h) P( A ∪ ( B ∩ C )) = 0.07 + 0.05 + 0.01 + 0.02 + 0.05 + 0.08 + 0.04 + 0.11 = 0.43
(i) P(( A ∪ B) ∩ C ) = 0.11 + 0.05 + 0.02 + 0.08 + 0.04 = 0.30
(j) P( A’ ∪ C ) = 0.04 + 0.03 + 0.05 + 0.08 + 0.06 + 0.13 + 0.11 + 0.11 + 0.07 + 0.02 +
0.05 + 0.08 + 0.04 = 0.87
P( A’ ∪ C)’ = 1 – P( A’ ∪ C) = 1 – 0.87 = 0.13
3. Suppose that a bag contains 43 red balls, 54 blue balls, and 72 green balls, and that 2
balls are chosen at random without replacement. Construct a probability tree for this
problem. What is the probability that 2 green balls will be chosen? What is the
probability that the 2 balls, chosen will have different colors?
SOLUTION:
Total number of balls:
43 red balls + 54 blue balls + 72 green balls = 169 balls
Probability Tree:
Let A be the event wherein 2 green balls are chosen
𝑃(𝐴) =
72
169
71
× 168 = 0.1801
Let B be the event wherein the 2 balls chosen have the same color,
And let C be the event wherein the 2 balls chosen will have different colors
𝑃(𝐶) = 1.0 – 𝑃(𝐵)
43
42
54
53
72
71
= 1.0 − [(169 × 168) + (169 × 168) + (169 × 168)]
= 1.0 − 0.3445
𝑃(𝐶) = 0.6556