Probability

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New answer posted

3 weeks ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

   

[as probability of getting odd number = probability of getting even number = ]

Probability of getting an odd number for odd number of times =

 

          

New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

General form Σ? ? C? (1/2)? < 1/2
Σ? ? C? < 2?
⇒? C? +? C? +.? C? < 128
⇒256− (? C? +? C? +.? C? ) < 128
⇒? C? +? C? +.? C? > 128
n≥5

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Given P (X=3) = 5P (X=4) and n=7
=>? C? p³q? = 5? C? p? q³
=> q = 5p and also p + q = 1
=> p = 1/6 and q = 5/6
Mean = 7/6 and variance = 35/36
Mean + Variance
= 7/6 + 35/36 = 77/36

New answer posted

3 weeks ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Total case = ¹? C?
Favourable cases
? C? (Select x? )
³C? (Select x? )
? C? (Select x? )
P = (6.3.7)/¹? C? = 1/68

New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Two drawn cards are spades. There are 50 cards left.
The missing card could be a spade or not a spade.
P (missing card is spade) = 11/50 (since 11 spades remain out of 50 cards).
P (missing card is not spade) = 1 - 11/50 = 39/50

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

P (x) = f (x³) + xg (x³) is divisible by x²+x+1. The roots of x²+x+1=0 are the complex cube roots of unity, ω and ω².
P (ω) = f (ω³) + ωg (ω³) = f (1) + ωg (1) = 0 — (I)
P (ω²) = f (ω²)³) + ω²g (ω²)³) = f (1) + ω²g (1) = 0 — (II)
Subtracting (II) from (I): (ω - ω²)g (1) = 0. Since ω ≠ ω², we must have g (1) = 0.
Substituting g (1)=0 into (I) gives f (1) = 0.
We need to find P (1) = f (1³) + 1*g (1³) = f (1) + g (1).
P (1) = 0 + 0 = 0.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

This is a binomial probability problem with n=5. Let p be the probability of success and q be the probability of failure.
Given P (X=1) =? C? p¹q? = 5pq? = 0.4096 — (I)
Given P (X=2) =? C? p²q³ = 10p²q³ = 0.2048 — (II)
Divide (I) by (II): (5pq? ) / (10p²q³) = 0.4096 / 0.2048 = 2.
(1/2) * (q/p) = 2 ⇒ q/p = 4 ⇒ q = 4p.
Using p+q=1, we have p+4p=1 ⇒ 5p=1 ⇒ p=1/5.
And q = 4/5.
We need to find P (X=3) =? C? p³q².
P (X=3) = 10 * (1/5)³ * (4/5)² = 10 * (1/125) * (16/25) = 160 / 3125 = 32 / 625.

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

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