2. According to a survey taken by an agency in a rural area, it has been observed that 75% of population treats diseases through self-medication without consulting a physician. Among the 12 residents surveyed on a particular day, find the probability that, (a) At least two of them treat diseases through self-medication without consulting a physician. (b) Exactly 10 of them consults physician before taking medication. (c) None of them consults physician before taking medication. (d) Less than 10 residents consult physician before taking medication. (e) All of them treat diseases through self-medication without consulting a physician.

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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2. According to a survey taken by an agency in a rural area, it has been observed that 75% of population
treats diseases through self-medication without consulting a physician. Among the 12 residents
surveyed on a particular day, find the probability that,
(a) At least two of them treat diseases through self-medication without consulting a physician.
(b) Exactly 10 of them consults physician before taking medication.
(c) None of them consults physician before taking medication.
(d) Less than 10 residents consult physician before taking medication.
(e) All of them treat diseases through self-medication without consulting a physician.
Transcribed Image Text:2. According to a survey taken by an agency in a rural area, it has been observed that 75% of population treats diseases through self-medication without consulting a physician. Among the 12 residents surveyed on a particular day, find the probability that, (a) At least two of them treat diseases through self-medication without consulting a physician. (b) Exactly 10 of them consults physician before taking medication. (c) None of them consults physician before taking medication. (d) Less than 10 residents consult physician before taking medication. (e) All of them treat diseases through self-medication without consulting a physician.
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Follow-up Questions
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Follow-up Question

I can see that you are changing the value of p and q depending on the question , is it ok to change them based on the question even though they are at the end from the same question just a different branch

 

in a. P was 0.75

in d . P was 0.25

↑
P(x-x) = S(12²) (0-75)* (0.25)/2-X
Lo
Let
x~ bin (n=12, p=0.75)
pmf of x
a) P(x>2) = 1- P(X<2)
= 1- [P(x=0] +P(X=1)]
Yes, you can
=
this method also
use
--[Rc (0-15) (0+ 167¹ +1 (0-153(1-0155]
[126 (-0.75
12G
= 1- [(0.25) ¹²¹ + 12 (0-75) (0-25)" ]
=defeat
| P(x=2) = 0-9999978(
Hoxe x~ Bin (n=12, p=0·25)
P(x=x) = [(1/²) (025) * (0.75 ) ²-
or
P(x<10) = 1- P(x=10)
JX=0, J₁2, 12
JO.W.
P(x ≤ 10) = 0.999962
1x=0, J₁ 2₁ 12
10W.
= |- [P(X=10] +P(X=11] +P(x=12]]
= 1-[(1/2) (0251) (0 7514 (17)(0+25 1075 14+(12)(025 075)
Ext
=
Yes, you can do both method.
P(x < 10) = 1- P(x = 10)
= 1- [P(x=10) + P(X=11)+P(X=12)
-|- P(x = 10) - P(x= 11 ) - P(x=R)
=
Step 2
So you can do both methods, getting a same
answers from both types (methods).
So, do any one of them.
Your thinking is also correct
you do like
add them and then subtract them from one.
Transcribed Image Text:↑ P(x-x) = S(12²) (0-75)* (0.25)/2-X Lo Let x~ bin (n=12, p=0.75) pmf of x a) P(x>2) = 1- P(X<2) = 1- [P(x=0] +P(X=1)] Yes, you can = this method also use --[Rc (0-15) (0+ 167¹ +1 (0-153(1-0155] [126 (-0.75 12G = 1- [(0.25) ¹²¹ + 12 (0-75) (0-25)" ] =defeat | P(x=2) = 0-9999978( Hoxe x~ Bin (n=12, p=0·25) P(x=x) = [(1/²) (025) * (0.75 ) ²- or P(x<10) = 1- P(x=10) JX=0, J₁2, 12 JO.W. P(x ≤ 10) = 0.999962 1x=0, J₁ 2₁ 12 10W. = |- [P(X=10] +P(X=11] +P(x=12]] = 1-[(1/2) (0251) (0 7514 (17)(0+25 1075 14+(12)(025 075) Ext = Yes, you can do both method. P(x < 10) = 1- P(x = 10) = 1- [P(x=10) + P(X=11)+P(X=12) -|- P(x = 10) - P(x= 11 ) - P(x=R) = Step 2 So you can do both methods, getting a same answers from both types (methods). So, do any one of them. Your thinking is also correct you do like add them and then subtract them from one.
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Follow-up Question

Sorry but, I am confused to why you subtract p( x= 0 ) - p(x=1)

why not add them and then subtract them from one like this:

1- p( p(x=0) + p(x=1))

? Can you elaborate

A 5G III.
VVICII POLII.
wring p=0.75
Than,
(@)
X =
)
n=12
X₁ Binomial (n=12, p=0-75)
stequired probability !_!!
1-
number of people treat disease throug self medication
without consulting a physical..
P(x > 2) = 1- P(X<2)
= 1 -
P(x = 0) - P(x = 1)
±1 - 126 (0.75) (1-0.75)¹2 - 12c, (0.75)' (1-0.75) 2-1
Ⓒ stequirel
= 0.999 9978
Here,
we need to find probility for variable defined for
who consults physician :-
required probability:
P(x= 10)
(0.25/¹² 12 * (0.75) + (0.25)"
12C 10 (1-0-75)¹⁰ (0.75)
0.0000354
Ⓒ In this case of all 12
person
Before taking medication, required probability! -
P(x = None of them)
126₂ (0.75) ¹² (1-0-75-112-12
0.75-12
0.0316763
=
12-10
siequired probability:- (when p=1-0.75= 0.25)
P ( x < 10) = 1- P(x >10)
stequired probability!
= 0.999962
-
9:00
mcr
= 1- P(x=10) - P(X=11) - P(X=12)
"
=1- 12₁ (0:25)¹0 (0.75 )² - 126, (025) " (075) - 12₁₂ (0.25) ² (0.75)
have not consuted the physican
P ( X = all of them) = 126 (0.25) ⁰ (0.75) 12-0
= 0·0316763
part Ⓒ and part are same question orked in
©
bartleby.com
mi
r! (n-r)!
different
way
६६
Transcribed Image Text:A 5G III. VVICII POLII. wring p=0.75 Than, (@) X = ) n=12 X₁ Binomial (n=12, p=0-75) stequired probability !_!! 1- number of people treat disease throug self medication without consulting a physical.. P(x > 2) = 1- P(X<2) = 1 - P(x = 0) - P(x = 1) ±1 - 126 (0.75) (1-0.75)¹2 - 12c, (0.75)' (1-0.75) 2-1 Ⓒ stequirel = 0.999 9978 Here, we need to find probility for variable defined for who consults physician :- required probability: P(x= 10) (0.25/¹² 12 * (0.75) + (0.25)" 12C 10 (1-0-75)¹⁰ (0.75) 0.0000354 Ⓒ In this case of all 12 person Before taking medication, required probability! - P(x = None of them) 126₂ (0.75) ¹² (1-0-75-112-12 0.75-12 0.0316763 = 12-10 siequired probability:- (when p=1-0.75= 0.25) P ( x < 10) = 1- P(x >10) stequired probability! = 0.999962 - 9:00 mcr = 1- P(x=10) - P(X=11) - P(X=12) " =1- 12₁ (0:25)¹0 (0.75 )² - 126, (025) " (075) - 12₁₂ (0.25) ² (0.75) have not consuted the physican P ( X = all of them) = 126 (0.25) ⁰ (0.75) 12-0 = 0·0316763 part Ⓒ and part are same question orked in © bartleby.com mi r! (n-r)! different way ६६
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Follow-up Question

Why did you say p = 1-0.75 rather than p =0.75?

Solution
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Follow-up Question

Is this right? I didn't understand your solution in a can elaborate

Cie
2. p=0.75
8
(9) P(XX²)
de xto
• P. (X=0) =X)
• P(x=1)
=
P (X-72)
12
(1²)
$19 = 0.25
pise)
=
perfet
That
(12)
1-
x 0.75
1 X 1 X 0.25
2-145 x 10-
120 X (ose
Sieles
(P(X=0) + P(X= 1))
x 0.75
-6
12
X 0.12
1
O
(12-1)
X
x 10.25 (01 5X) CA
12
5.96X10-8
n=
11
10
x 0.25
53-5
= 0.999 299.9% 0
625
12
NOT
STY
1- (5.96× 10 -8 + 2+145 x 10-6)
1
10
0535 +
X
(b) Since the probabilty is exactly 10 of the 12 recidents
Consult the physican we can etter say say
q=0-45 Faiture
2103104
(0=x) ()
(1
2 reciclents don't
P don't consult: 0.750.09
00112-10
0.75
x (0.25)
(X=0
(X=10]
11=X) (
(01) *) q
92.0=1
@nsult
1677
Transcribed Image Text:Cie 2. p=0.75 8 (9) P(XX²) de xto • P. (X=0) =X) • P(x=1) = P (X-72) 12 (1²) $19 = 0.25 pise) = perfet That (12) 1- x 0.75 1 X 1 X 0.25 2-145 x 10- 120 X (ose Sieles (P(X=0) + P(X= 1)) x 0.75 -6 12 X 0.12 1 O (12-1) X x 10.25 (01 5X) CA 12 5.96X10-8 n= 11 10 x 0.25 53-5 = 0.999 299.9% 0 625 12 NOT STY 1- (5.96× 10 -8 + 2+145 x 10-6) 1 10 0535 + X (b) Since the probabilty is exactly 10 of the 12 recidents Consult the physican we can etter say say q=0-45 Faiture 2103104 (0=x) () (1 2 reciclents don't P don't consult: 0.750.09 00112-10 0.75 x (0.25) (X=0 (X=10] 11=X) ( (01) *) q 92.0=1 @nsult 1677
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