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

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*q^2n+1-0

+ ²ⁿ⁺¹C₁*p¹*q^2n+1-1
+²ⁿ⁺¹C₂*p²*q^2n+1-2

.....+

²ⁿ⁺¹C_n*pⁿ*q^2n+1-n

Since, p=q=0.5,

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

= 0.5²ⁿ⁺¹*[0.5*[²ⁿ⁺¹C₀+²ⁿ⁺¹C₁...²ⁿ⁺¹C_2n+1]]

= 0.5²ⁿ⁺¹*[0.5*2²ⁿ⁺¹]

=0.5

pls explain clearly d second last step of ur solution. i got struck at that step only

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

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.