Page 189 - DMTH404_STATISTICS

Unit 14: Theoretical Probability Distributions

Notes
2 25
5
1ö æ ö
1 2
(b) P (r = ) 1 = 3 C p q = 3. æ ç ÷ ç ÷ = 72
1
6 ø è ø
è
6
125 91
(c) Probability of getting at least one six = 1 - P(r = 0) 1=  =
216 216
2 5 5
1ö æ ö
æ
2 1
(d) P (r = ) 2 = 3 C p q = 3. =
2 ç ÷ ç ÷ 72
6 ø è ø
è
6
3 1
1ö
3 0
(e) P (r = ) 3 = 3 C p q = 3. æ ç ÷ =
3
è 6 ø 216
Example 2:
Assuming that it is true that 2 in 10 industrial accidents are due to fatigue, find the probability
that:
(a) Exactly 2 of 8 industrial accidents will be due to fatigue.
(b) At least 2 of the 8 industrial accidents will be due to fatigue.
Solution.

Eight industrial accidents can be regarded as Bernoulli trials each with probability of success
2 1
p = = . The random variable r denotes the number of accidents due to fatigue.
10 5

2 4 6
1ö æ ö
æ
(a) P (r = ) 2 = 8 C 2 ç ÷ ç ÷ = 0.294
è 5 ø è ø
5
(b) We have to find P(r  2). We can write
P(r  2) = 1 – P(0) – P(1), thus, we first find P(0) and P(1).

0 4 8
æ
1ö æ ö
We have P ( ) 0 = 8 C 0 ç ÷ ç ÷ = 0.168
5 ø è ø
è
5
1 4 7
æ
1ö æ ö
and P ( ) 1 = 8 C 1 ç ÷ ç ÷ = 0.336
è
5 ø è ø
5
 P(r  2) = 1 – 0.168 – 0.336 = 0.496
Example 3: The proportion of male and female students in a class is found to be 1 : 2.
What is the probability that out of 4 students selected at random with replacement, 2 or more
will be females?
Solution.

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