Question

if a die is cast untill 6 appears then the probability that it must be cast more than five times is​

Answer 1

Answer:

Step-by-step explanation:

According to the question given, we have to throw a dice until 6 appears.

And we have to find the probability that the dice is thrown more than 5 times.

i.e. 6 must not appear in the Ist, 2nd, 3rd, 4th, and 5th throw.

It would be easy to solve this question by negation. i.e. finding the probability that 6 appears either in 1st , 2nd, 3rd, 4th, or 5th throw. And then Subtracting this probability by 1, we will get our probability.

Case 1: Probability of getting 6 in 1st throw:-

P(1st) = 1/6

Case 2: Probability of getting 6 in 2nd throw:-

P(2nd) = 5/6 * 1/6

Case 3: Probability of getting 6 in 3rd throw:-

P(3rd) = 5/6 * 5/6 * 1/6

Case 4: Probability of getting 6 in 4th throw:-

P(4th) = 5/6 * 5/6 * 5/6 * 1/6

Case 5: Probability of getting 6 in 5th throw:-

P(5th) = 5/6 * 5/6 * 5/6 * 5/6 * 1/6.

Now P(More than 5 times) = 1 - {P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th)}

P(More than 5 times) = 1 - { 1/6 + 5/6*1/6 + (5/6)^2 *1/6 + (5/6)^3 *1/6 + (5/6)^4 *1/6}

P(More than 5 times) = 1 - 1/6{1 + 5/6 + (5/6)^2 + (5/6)^3 + (5/6)^4}

P(More than 5 times) = 1 - 1/6{(1 - (5/6)^5)/(1 - 5/6)} [Sum of n terms of a GP series.]

P(More than 5 times) = 1 - {1 - (5/6)^5}

P(More than 5 times) = 1 - 1 + (5/6)^5

P(More than 5 times) = (5/6)^5

Hence the required probability is (5/6)^5.