Page 278 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 13: Binomial Probability Distribution

r
n
1 n r
r
or C p q n r n C p q 1 Notes
r r 1
! n r n r ! n r 1 n r 1
or p q p q
! r n r ! r 1 ! n r 1 !
1 1
p
or .q . or qr q np pr
n r r 1
Solving the above inequality for r, we get
r n 1 p 1 .... (1)
Similarly, on solving the inequality P(r - 1) P(r) for r, we can get
r n 1 p .... (2)

Combining inequalities (1) and (2), we get
n 1 p 1 r n 1 p

Case I. When (n + 1)p is not an integer
When (n + 1)p is not an integer, then (n + 1)p - 1 is also not an integer. Therefore, mode will
be an integer between (n + 1)p - 1 and (n + 1)p or mode will be an integral part of (n + 1)p.
Case II. When (n + 1)p is an integer
When (n + 1)p is an integer, the distribution will be bimodal and the two modal values
would be (n + 1)p - 1 and (n + 1)p.

Example: An unbiased die is tossed three times. Find the probability of obtaining (1) no
six, (2) one six, (3) at least one six, (4) two sixes and (5) three sixes.
Solution.
The three tosses of a die can be taken as three repeated trials which are independent. Let the
occurrence of six be termed as a success. Therefore, r will denote the number of six obtained.
1
Further, n = 3 and p .
6
1. Probability of obtaining no six, i.e.,
0 5 3 125
1
0 3
P r 0 3 C p q 1.
0
6 6 216
2 25
1
5
1 2
2. P r 1 3 C p q 3.
1
6 6 72
125 91
3. Probability of getting at least one six = 1 – P(r = 0) 1
216 216
2
1 5 5
2 1
3
4. P r 2 C p q 3.
2
6 6 72
3 1
1
3 0
5. P r 3 3 C p q 3.
3
6 216
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