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Q1) a coin is tossed n times. What is the chance that the head will present itself odd number of time? A)2^n/n-1 B)2^n/n C)1/2 D)1/2^n Q2)There are 4 addressed envelops and 4 letters. Then the probability that all the letters are not mailed through proper envelop is: A)1/24 B)1 C)23/24 D)3/4

Q1) a coin is tossed n times. What is the chance that the head will present itself odd number of time?


A)2^n/n-1
B)2^n/n
C)1/2
D)1/2^n

Q2)There are 4 addressed envelops and 4 letters. Then the probability that all the letters are not mailed through proper envelop is:
A)1/24
B)1
C)23/24
D)3/4

Grade:12

4 Answers

Pratham Ashish
17 Points
13 years ago

Dear Varsha ,

2nd ques is quite straight forward as

we can send  the letters and envelopes in 4! different ways out of which there will be only one way that letters are mailed properly wid der envelopes then

according to d quest, all letters are not mailed properly which means total ways minus one way of proper mail,

p = (4! - 1)/ 4! = 23/24

ans = (c)

 

Pratham Ashish
17 Points
13 years ago

Dear Varsha ,

Q 1

if d coin is tossed n times ,den total possibilities will be 2^n.

let chance of getting r head out of n be nCr

den total way of getting odd times head will be nC1 + nC3+ nC5 +.......nCn (if n is odd) 0r nC1 + nC3+ nC5 +.......nC(n-1) ,for n is even

nw, (1 + x)^n = nC0+ nC1 x + nC2 x^2+........

put x=1 , 2^n = 1 + nC1 + nC2 +......

put x = -1, 0 = 1 - nC1 + nC2 - nC3 +......

just take difference of d above 2 eqn ,

2^n = 2( nC1 + nC3 +........)

if n is odd den

nC1 + nC3+ nC5 +.......nCn = 2^(n-1), probability,p = 2^(n-1)/2^n = 1/2.

 

Pratham Ashish
17 Points
13 years ago

hi varsha ,

Q2

toatal no of ways in which 4 letters n  4 envelopes can be arranged is 4!

der can be only 1 way in which in which all letters are mailed properly wid der envelopes.

so required probability,p = (4!-1)/4!   [ all d letters not mailed properly = total ways - no of way in which dey r mailed properly ]

p = 23/24

Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

(1) if d coin is tossed n times ,den total possibilities will be 2^n.
let chance of getting r head out of n be nCr
den total way of getting odd times head will be nC1 + nC3+ nC5 +.......nCn (if n is odd) 0r nC1 + nC3+ nC5 +.......nC(n-1) ,for n is even
nw, (1 + x)^n = nC0+ nC1 x + nC2 x^2+........
put x=1 , 2^n = 1 + nC1 + nC2 +......
put x = -1, 0 = 1 - nC1 + nC2 - nC3 +......
just take difference of d above 2 eqn ,
2^n = 2( nC1 + nC3 +........)
if n is odd den
nC1 + nC3+ nC5 +.......nCn = 2^(n-1), probability,p = 2^(n-1)/2^n = 1/2.

(ii) we can send the letters and envelopes in 4! different ways out of which there will be only one way that letters are mailed properly wid der envelopes then
according to d quest, all letters are not mailed properly which means total ways minus one way of proper mail,
p = (4! - 1)/ 4! = 23/24
ans = (c)

Thanks and Regards

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