Two dice are thrown n times in succession. What is the probability of obtaining ‘double six’at least once? Find the least number of throws so that the probability of obtaining at least one‘double six’ will be greater than 1/2?

Sure, here is the step by step solution:

Step 1: Understand the problem We are throwing two dice together and we want to find the probability of getting a 'double six' at least once. A 'double six' means that both dice show a six.

Step 2: Calculate the probability of getting a 'double six' in one throw There are 6*6 = 36 possible outcomes when two dice are thrown. Only one of these outcomes is a 'double six'. So, the probability of getting a 'double six' in one throw is 1/36.

Step 3: Calculate the probability of not getting a 'double six' in one throw The probability of not getting a 'double six' in one throw is 1 - 1/36 = 35/36.

Step 4: Calculate the probability of not getting a 'double six' in n throws The probability of not getting a 'double six' in n throws is (35/36)^n.

Step 5: Calculate the probability of getting a 'double six' at least once in n throws The probability of getting a 'double six' at least once in n throws is 1 - (35/36)^n.

Step 6: Find the least number of throws so that the probability of getting at least one 'double six' is greater than 1/2 We need to solve the inequality 1 - (35/36)^n > 1/2. Solving this inequality gives n > log(2)/log(36/35) ≈ 24.6. Since n must be an integer, the least number of throws is 25.