Probability

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a month ago

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V
Vishal Baghel

Contributor-Level 10

n (s) = 6² = 36
E = { (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), (2, 1), (2, 2), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 5), (5, 1), (5, 2), (5, 3), (7, 1)}
∴ n (E) = 17
Required prob. = n (E) / n (S) = 17 / 36

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

11.00
Let probability of hitting the target = p ⇒ p=1/2
Let n be the minimum number of bombs
According to given condition
1 - (?C?P?(1-P)? + ?C?P¹(1-P)?¹) ≥ 99/100
⇒ 2? ≥ (n+1)100
n=10 ⇒ 2¹? ≥ 1100 Reject
n=11 ⇒ 2¹¹ ≥ 1200 Select

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Given, mean = np = . and variance = npq = α3q=13andp=23

P (X=1)=np1qn1=4243n (23)1 (13)n1=4243n=6

P (X=4or5)=6C4 (23)4 (13)2+6C5 (23)5 (13)1=1627

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

Mean = 4 = μ = np

Variance = σ2=np (1p)=434 (1p)=43p=23n=6

=6C0 (13)6+6C1 (23)1 (13)6+6C2 (23)2 (13)4=14627

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

X  0 1 2 3

P (X) 16 12 310 130

σ2=Σx2P (x) (Σ*P (x)2=56100)

100σ2=56

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

 P (A/B)=17P (AB)P (B)=17

P (B)=79

P (B/A)=25P (AB)P (A)=25

P (A)=518

Now,  P (A'B)=1P (AB)+P (B)

Both (S1 ) and (S2 ) are true

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

(2x3+3xk)12

gen term =

=12Cr212r.3r.x363rrk

For constant term

36 – 3r – rk = 0

k=363rr

for r = 1, 2, 4

12Cr212-r>28

Possible values of k = 3, 1

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

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New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

P (En) = n/36 for n = 1, 2, 3, …., 8

P (A)=Anypossiblesumof (1, 2, 3, ........., 8) (=αsay)36


α3645

a29

If one of the number from {1, 2, ….8} is left then total  29 by 3 ways

Similarly by leaving terms more 2 or 3 we get 16 more combinations

 Total number of different set a possible is 16 + 3

= 19

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