Page 190 - DMTH404_STATISTICS

Statistics

Notes Let the selection of a female student be termed as a success. Since the selection of a student is
made with replacement, the selection of 4 students can be taken as 4 repeated trials each with
2
probability of success p = .
3

Thus, P(r  2) = P(r = 2) + P(r = 3) +P(r = 4)

2 2 3 4
1
1
4 æ 2ö æ ö 4 æ 2ö æ ö 4 æ 2ö 8
= C 2 ç ÷ ç ÷ + C 3 ç ÷ ç ÷ + C 4 ç ÷ = 9
è
3 ø è ø
è
è
3ø
3 ø è ø
3
3
Note that P(r  2) can alternatively be found as 1 - P(0) - P(1)
Example 4: The probability of a bomb hitting a target is 1/5. Two bombs are enough to
destroy a bridge. If six bombs are aimed at the bridge, find the probability that the bridge is
destroyed.
Solution.
1
Here n = 6 and p =
5
The bridge will be destroyed if at least two bomb hit it. Thus, we have to find P(r ³ 2). This is
given by

6 5
4
6 æ 4ö 6 æ 1ö æ ö 1077
P(r  2) = 1 – P(0) – P(1) 1=  C 0 ç ÷  C 1 ç ÷ ç ÷ =
è 5 ø è 5 ø è ø 3125
5
Example 5: An insurance salesman sells policies to 5 men all of identical age and good
health. According to the actuarial tables, the probability that a man of this particular age will be
alive 30 years hence is 2/3. Find the probability that 30 years hence (i) at least 1 man will be
alive, (ii) at least 3 men will be alive.
Solution.
Let the event that a man will be alive 30 years hence be termed as a success. Therefore, n = 5 and
2
p = .
3
0 5
F I F I 1 242
2
5
(i) P(r ³ 1) = 1 - P(r = 0) =  G J G J =1 C 0
H K H K 3 243
3
(ii) P(r ³ 3) = P(r = 3) + P(r = 4) +P(r = 5)
5 F F 1I 2 5 F F 1I 5 F 2I 5 64
2I
2I
4
3
= C 3 3 H K H K + C 4 3 H K H K + C 5 3 H K = 81
3
3
Example 6: Ten percent of items produced on a machine are usually found to be defective.
What is the probability that in a random sample of 12 items (i) none, (ii) one, (iii) two, (iv) at the
most two, (v) at least two items are found to be defective?
Solution.

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