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a dice is thrown (2n+1) times.the probability of getting 1,3 or 4 at most n times is:

a dice is thrown (2n+1) times.the probability of getting 1,3 or 4 at most n times is:

Grade:12

4 Answers

askiitian expert akshay singh
8 Points
11 years ago

p=probabilty of getiing 1/3/4 in one throw

q=1-p

P( getting 1,3 or 4 at most n times)=P(0 times) +P(1 time)....P(n times)

=2n+1C0*p0*q2n+1-0

+ 2n+1C1*p1*q2n+1-1
+2n+1C2*p2*q2n+1-2

.....+

2n+1Cn*pn*q2n+1-n

Since, p=q=0.5,

so above = 0.52n+1*[2n+1C0+2n+1C1...2n+1Cn]

= 0.52n+1*[0.5*[2n+1C0+2n+1C1...2n+1C2n+1]]

= 0.52n+1*[0.5*22n+1]

=0.5

 

varsha jayant karandikar
18 Points
11 years ago
pls explain clearly d second last step of ur solution. i got struck at that step only
mohit yadav
24 Points
11 years ago

hi varsha,

u have a basic method given by akshay sir u can also do dis question in an objective exam by putting n=0 so d question bcomes throwing d die 1 time and we wanna get d probability of getting no 1,3or4 so probability=3/6=0.5 or do by taking n=1 but for subjective do by sir's method

askiitian expert akshay singh
8 Points
11 years ago

hi varsha ,

  the last step was 0.52n+1*[0.5*22n+1].........now ,  0.52n+1 =(1/2)2n+1  so 22n+1 within then bracket cancels out with (1/2)2n+1 outside the bracket ,so the answer is 0.5.

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