The probability that a prime number occurs on at least 2 tosses when a balanced dice is tossed 5 times independently is equal to???

To solve this probability problem, we need to determine the probability of getting a prime number on each individual dice toss, and then calculate the probability of getting at least 2 prime numbers out of 5 tosses.

Step 1: Determine the probability of getting a prime number on a single dice toss.
A balanced dice has 6 sides numbered from 1 to 6. To find the probability of getting a prime number, we need to determine how many prime numbers are on the dice. The prime numbers between 1 and 6 are 2, 3, and 5. So, the probability of getting a prime number on a single toss is 3/6 or 1/2.

Step 2: Calculate the probability of getting at least 2 prime numbers out of 5 tosses.
To calculate this probability, we will consider all the possible combinations of getting prime and non-prime numbers on the 5 tosses.

There are different ways to approach this calculation, but one simple method is to use the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)

Where:
P(X = k) is the probability of getting exactly k successes,
(n choose k) is the number of ways to choose k successes out of n trials,
p is the probability of a single success, and
(1 - p) is the probability of failure or getting a non-prime number.

To find the probability of getting at least 2 prime numbers, we sum up the probabilities of getting exactly 2, 3, 4, and 5 prime numbers.

P(at least 2 prime numbers) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Let's calculate this step by step:
P(X = 2) = (5 choose 2) * (1/2)^2 * (1 - 1/2)^(5 - 2)
= (10) * (1/4) * (1/2)^3
= 10/32
= 5/16

P(X = 3) = (5 choose 3) * (1/2)^3 * (1 - 1/2)^(5 - 3)
= (10) * (1/8) * (1/2)^2
= 10/64
= 5/32

P(X = 4) = (5 choose 4) * (1/2)^4 * (1 - 1/2)^(5 - 4)
= (5) * (1/16) * (1/2)^1
= 5/32

P(X = 5) = (5 choose 5) * (1/2)^5 * (1 - 1/2)^(5 - 5)
= (1) * (1/32) * (1/2)^0
= 1/32

Finally, we can calculate the probability of getting at least 2 prime numbers:
P(at least 2 prime numbers) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 5/16 + 5/32 + 5/32 + 1/32
= 16/32 + 5/32 + 5/32 + 1/32
= 27/32

Therefore, the probability that a prime number occurs on at least 2 tosses when a balanced dice is tossed 5 times independently is 27/32.