Probstat/week2 practice 1 solutions - ประวัติรุ่นแก้ไข

Jittat เมื่อ 06:45, 13 ตุลาคม 2557

2014-10-13T06:45:09Z

Jittat: /* Conditional probability review */

2014-10-13T06:36:21Z

Conditional probability review

Jittat: หน้าที่ถูกสร้างด้วย 'This are solutions to Probstat/week2 practice 1. == Conditional probability review == 1.1 There are 7 numbers divisible by 2 and ...'

2014-10-13T06:32:01Z

หน้าที่ถูกสร้างด้วย 'This are solutions to Probstat/week2 practice 1. == Conditional probability review == 1.1 There are 7 numbers divisible by 2 and ...'

หน้าใหม่

This are solutions to [[Probstat/week2 practice 1]].

== Conditional probability review ==

1.1 There are 7 numbers divisible by 2 and 3 (0, 6, 12, 18, 24, 30, 36). There are 14 numbers divisible by 3. Let '''A''' be the event that a random number is divisible by 3. Let '''B''' be the event that a random number is divisible by 2. We want to compute '''P(B|A)''', which is '''P(AB)/P(A)''' = 7/14 = 1/2.

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	1.4 Note that '''P(C)''' = 11/42. This is not equal to '''P(C\|A)'''. Therefore both events are not independent.		1.4 Note that '''P(C)''' = 11/42. This is not equal to '''P(C\|A)'''. Therefore both events are not independent.
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		+	2.1 '''P(C\|S) = P(CS)/P(S)''' = 0.12 / (0.12+0.03) = 12/15 = 4/5.
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		+	2.2 '''P(C<sup>c</sup>\|S<sup>c</sup>) = P(C<sup>c</sup>S<sup>c</sup>)/P(S<sup>c</sup>)''' = 0.78 / (0.78+0.07) = 0.78/0.85. (Approximately 0.91764)
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		+	2.3 '''P(S\|C) = P(SC)/P(C)''' = 0.12 / (0.12 + 0.07) = 0.12/0.19 (approximately 0.6316)
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		+	2.4 It predicts incorrectly when the person smokes and does not have cancer or when the person does not smoke but has cancer. The probability is '''P(SC<sup>c</sup> U S<sup>c</sup>C) = P(SC<sup>c</sup>) + P(S<sup>c</sup>C)''' = 0.07 + 0.03 = 0.1.
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		+	== Counting ==

←รุ่นแก้ไขก่อนหน้า		รุ่นแก้ไขเมื่อ 06:36, 13 ตุลาคม 2557
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	1.1 There are 7 numbers divisible by 2 and 3 (0, 6, 12, 18, 24, 30, 36). There are 14 numbers divisible by 3. Let '''A''' be the event that a random number is divisible by 3. Let '''B''' be the event that a random number is divisible by 2. We want to compute '''P(B\|A)''', which is '''P(AB)/P(A)''' = 7/14 = 1/2.		1.1 There are 7 numbers divisible by 2 and 3 (0, 6, 12, 18, 24, 30, 36). There are 14 numbers divisible by 3. Let '''A''' be the event that a random number is divisible by 3. Let '''B''' be the event that a random number is divisible by 2. We want to compute '''P(B\|A)''', which is '''P(AB)/P(A)''' = 7/14 = 1/2.
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		+	1.2 Using the definition of events '''A'''and '''B''' from question 1.1, note that '''P(B) = 21/42 = 1/2'''. Thus '''P(B\|A) = P(B)'''. This implies that '''P(AB)=P(A)P(B)'''. Therefore, both events are independent.
		+
		+	1.3 The number divisible by 4 and 3 are 0, 12, 24, and 36. Let '''C''' be the event that a random number is divisible by 4. Thus, '''P(C\|A)''' = 4/14 = 1/7.
		+
		+	1.4 Note that '''P(C)''' = 11/42. This is not equal to '''P(C\|A)'''. Therefore both events are not independent.