Probability

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a month ago

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A
alok kumar singh

Contributor-Level 10

Total = 9 (10? )
Fav. Way =? C? (2? -2) +? C? (2? -1) = 36 (30) + 9 (15) = 1080 + 135
Probability = (36x30+9x15)/ (9x10? ) = (4x30+15)/10? = 135/10?

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Sum obtained is a multiple of 4.
A = { (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3) (6,2), (6,6)}
B: Score of 4 has appeared at least once.
B = { (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6)}
Required probability = P (B/A) = P (B? A)/P (A)
= (1/36) / (9/36) = 1/9

New question posted

a month ago

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New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly considere the following Image 

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Let B? be the event where Box-I is selected and B? →where box-II selected
P (B? )=P (B? )=1/2
Let E be the event where selected card is non prime.
For B? : Prime numbers: {2,3,5,7,11,13,17,19,23,29}
For B? : Prime numbers: {31,37,41,43,47}
P (E)=P (B? )*P (E/B? )+P (B? )P (E/B? )
= 1/2*20/30+1/2*15/20
Required probability:
P (B? /E) = (P (E/B? )P (B? )/P (E) = (1/2*20/30)/ (1/2*20/30+1/2*15/20) = (2/3)/ (2/3+3/4) = 8/17

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

MD²+MC² is minimum when M is the midpoint of projection of CD on AB.

 

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