This what I have so far but I need help on 1(d) and number 3 a and b 1. 1. An ur
ID: 3208210 • Letter: T
Question
This what I have so far but I need help on 1(d) and number 3 a and b
1. 1. An urn contains 4 red balls, 3 blue balls, 2 green balls and one black ball. One ball is selected at random. Consider games X defined as follows.
Red
Blue
Green
Black
X
-$5.00
+$1.00
+$2.50
+10.00
(a) Find the probabilities P(X > 1.00) and P(X 1.00).
P(x>1) = P(green) + P(black) = 2/10 + 1/10 = 3/10
P(x>=1) = 3/10 + p(blue) = 3/10 + 3/10 = 2/5
(b) Find the expected value, E(X).
E(x) = sum of (X x P(X)) = (-5x5/10) + (1x3/10) + (2.5x2/10) + (10x1/10)
= -7/10 = -0.7
(c) Find the probability standard deviation, (X).
variance = (-5+0.7)2x0.5 + (1+0.7)2x0.3+(2.5+0.7)2x0.2 + (10+0.7)2 x 0.1
= 23.61
Standard deviation = square root (23.61) = 4.86
(d) Find the cumulative distribution functions F(x) = P(X x)
2. A cumulative distribution function of a random variable X is given as follows.
F(x) = 0.00, if x < 2.0,
0.30, if 2.0 x < 1.5,
0.85, if 1.5 x < 2.0,
0.90, if 2.0 x < 3.25,
1.00, if 3.25 x.
(a) Find the probability mass function..
X is defined at the points -2, 1.5, 2.0, 3.25. So the probability mass function of X will be
P(X=-2) = 0.30
P(X=1.5) = 0.85 p(x<1.5) = 0.85-0.30 =0.55
P(X=2.0) = 0.90 p(x<2.0) = 0.90-0.85 =0.05
P(X=3.25) = 1.0 p(x<3.25) = 1-0.90 =0.10
(b) Find the probability, P(1.0 X 2.0).
The probability P(1.0 X 2.0) will be
0.60
(c) Find the expected value.
X
P(X=x)
XP(X=x)
X^2P(X=x)
-2
0.3
-0.6
1.2
1.5
0.55
0.825
1.2375
2
0.05
0.1
0.2
3.25
0.1
0.325
1.05625
Total
0.65
3.69375
0.60
(d) Find the probability standard deviation.
1.8087
3. From the urn of Problem 1, ten balls are selected at random with replacement.
(a) Estimate the probability that four or less red balls are selected.
(b) Estimate the probability that more than four blue balls are selected.
Red
Blue
Green
Black
X
-$5.00
+$1.00
+$2.50
+10.00
Explanation / Answer
3a) the probability that four or less red balls are selected
(Given the case of picking 10 balls with replacement, i.e one ball is picked and put back in urn and next one ball is picked...event ends when 10 balls are picked and put them back in bag)
Assuming the 4 red balls are identical
4red balls = (4/10)^4 * (6/10)^6
3red balls = (4/10)^3 * (6/10)^7
2red balls = (4/10)^2 * (6/10)^8
1 red ball = (4/10)^1 * (6/10)^9
0 red balls = (6/10)^10
Add all the above cases = 0.0015
3 b) probability that more than four blue balls are selected
= 0.3 ^ 5 * 0.7 ^ 5 + 0.3 ^6 * 0.7^4 +....... 0.3^10
= 0.0007
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