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Unformatted text preview: obability that all 3 are found nondefective.
Therefore the probability that the lot is rejected is 1P(accept the lot). P (accept) = 0.30 46
19
0
3
10 + 0.70 0 10 = 0.54.
3
3 3 So, P(reject)= 1−P(accept)=10.54=0.46.
Example 3
This is geometric problem with p =
a. P (X = n) = ( M
M +N . N
M
M N n−1
)n−1
=
.
N +M
N +M
( M + N )n b. P (X ≥ k ) = ∞
x=k ( N
M
)x−1
N +M
N +M M
M +N ∞
x=k ( = N
)x−1
M +N = M
N
N
N
[(
)k−1 + (
)k + · · · ] = (
)k−1 .
M +N M +N
M +N
M +N
Example 4
This is a binomial problem together with some probability (law of total probability and Bayes’ theorem).
Let Li be the event that lot was produced on line i.
Let X be the number of defective fuses among the 3 selected.
a.
P (L1 ∩ X = 1)
P (X = 1) = P (X = 1/L1 )P (L1 )
P (X = 1/L1 )P (L1 ) + P (X = 1/L2 )P (L2 ) + P (X = 1/L3 )P (L3 ) + P (X = 1/L4 )P (L4 ) + P (X = 1/L5 )P (L5 ) = P (L1 /X = 1) = 3
1 0.051 0.952 1 +
5 3
1 3
0.051 10.952 1
1
5
3
3
1 0.021 0.982 1 +
5 1 0.021 0.982 5 + b. The probability that the lot came from one of the other 4 lines is:
P (L2 ∪ L3 ∪ L4 ∪ L5 /X = 1) = 1 − P (L1 /X = 1) = 1 − 0.37 = 0.63. 15 1 0.021 0.982 1 +
5 3
1 0.021 0.982 1
5 P (L1 /X = 1) = 0.37. ⇒ Example 5
Let X be the number of tests needed for each group of 10 people. Then, if nobody has the disease 1 test is enough. But if the
test is positive then there will be 11 test (1 + 10). The probability distribution of X is:
X
P (X )
10
1
0.100 0.9010 = 0.9010
0
11
1 − 10 0.100 0.9010 = 1 − 0.9010
0
Therefore the expected number of tests is:
10 + 11(1 − 0.9010 = 7.51.
E (X ) = 1(0.90)
Example 6
This is a hypergeometric problem. Let’s deﬁne the events:
D: Selected card is ♦ (part a)
H 4: All 4 selected cards are ♥ (part b)
Lc : Lost card is ♣
Ls : Lost card is ♠
Lh : Lost card is ♥
Ld : Lost card is ♦
a. We need to ﬁnd P (Lc /D). This equal to:
P (Lc /D) = P (Lc ∩ D)
P (D ) P (Lc ∩ D)
P (Lc ∩ D) + P (Ls ∩ D) + P (Lh ∩ D) + P (Ld ∩ D)
P (D/Lc )P (Lc )
P (D/Lc )P (Lc ) + P (D/Ls )P (Ls ) + P (D/Lh )P (Lh ) + P (D/Ld )P (Ld )
13 1
51 4 + 13 1
51 4
13 1
+ 13 1
51 4
51 4 + 12 1
51 4 = =
=
= 13
= 0.2549.
51 b. We need to ﬁnd P (Lc /H 4). This equal to:
P (Lc /H 4) = P (Lc ∩ H 4)
P (H 4) P (Lc ∩ H 4)
P (Lc ∩ H 4) + P (Ls ∩ H 4) + P (Lh ∩ H 4) + P (Ld ∩ H 4)
P (H 4/Lc )P (Lc )
P (H 4/Lc )P (Lc ) + P (H 4/Ls )P (Ls ) + P (H 4/Lh )P (Lh ) + P (H 4/Ld )P (Ld ) 1338
4
510 1
4
4
1338
1338
1239
1338
4
510 1 + 4510 1 + 4510 1 + 4510 1
4
4
4
4
4 4 13
4 4 + 13
13 4 12
4 + 4 4 + 13 = 0.2708.
4 Example 7
Here, X takes on two values 0 or 1. Let P (X = 1) = p. The probability distribution of X is:
X
P (X )
0
1p
1
p
a. E (X ) = 0(1 − p) + 1(p) ⇒ E (X ) = p.
V ar(X ) = 02 (1 − p) + 12 (p) − P 2 ⇒ V ar(X )...
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This note was uploaded on 10/04/2012 for the course STATISTICS 100a taught by Professor Cristou during the Spring '10 term at UCLA.
 Spring '10
 CRISTOU
 Binomial

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