The document contains two examples involving binomial distributions: 1) The probability of a component surviving a shock test is 3/4. The question asks for the probability that exactly 2 of the next 4 components will survive. 2) The probability that a patient recovers from a rare blood disease is 0.4. The question asks for the probabilities of: a) at least 10 of 15 people surviving, b) between 3 to 8 of 15 people surviving, and c) exactly 5 of 15 people surviving.

1. President University Erwin Sitompul PBST 6/1
Binomial Distribution
Chapter 5.3 Binomial and Multinomial Distributions
The probability that a certain kind of component will survive a given
shock test is 3/4. Find the probability that exactly 2 of the next 4
components tested will survive.
3
4
p 
2 2
4 2
3 3 1
2: 4,
4 4 4
b C
     
     
     
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

2. President University Erwin Sitompul PBST 6/2
Binomial and Multinomial Distributions
Chapter 5.3 Binomial and Multinomial Distributions
The probability that a patient recovers from a rare blood disease is
0.4. If 15 people are known to have contracted this disease, what is
the probability that (a)at least 10 survive, (b)from 3 to 8 survive,
and (c)exactly 5 survive?
Let X be the number of people that survive. Table A.1 gives help.
(a)
(b)
( 10) 1 ( 10)P X P X   
9
0
1 ( ;15,0.4)
x
b x

   1 0.9662  0.0338
15
10
( ;15,0.4)
x
b x


8
3
(3 8) ( ;15,0.4)
x
P X b x

   
8 2
0 0
( ;15,0.4) ( ;15,0.4)
x x
b x b x
 
  
0.9050 0.0271  0.8779
(c) ( 5) (5;15,0.4)P X b  0.18595 10
15 5 (0.4) (0.6)C
?Can you calculate
manually?