A die is rolled 6 times. Find the probability P(of exactly 4 occurrences of 2).
prob(2) = 1/6
prob(not 2) = 5/6
prob (four 2's out of 6)
= C(6,4) (1/6)^4 (5/6)^2
= 15(1/1296)(25/36)
= appr .008
To find the probability of exactly 4 occurrences of the number 2 when rolling a die 6 times, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
Let's break this down step by step:
1. Determine the total number of possible outcomes when rolling a die 6 times:
Since each roll of a die has 6 possible outcomes (numbers 1, 2, 3, 4, 5, and 6), the total number of possible outcomes for rolling a die 6 times is 6^6 (6 raised to the power of 6) because each roll is independent of the others.
So, the total number of possible outcomes = 6^6 = 46656.
2. Calculate the number of favorable outcomes:
To find the number of favorable outcomes, we need to count how many ways we can roll exactly 4 occurrences of the number 2 out of 6 rolls.
We can use the combinatorial concept of "n choose k" to calculate this. The formula for "n choose k" is:
C(n, k) = n! / (k!(n-k)!)
In this case, n = 6 (number of rolls) and k = 4 (number of occurrences of 2).
Plugging these into the formula, we get:
C(6, 4) = 6! / (4!(6-4)!) = 6! / (4! x 2!) = (6 x 5 x 4 x 3) / (4 x 3 x 2 x 1) = 90.
So, the number of favorable outcomes = 90.
3. Calculate the probability:
Finally, we can find the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
P(of exactly 4 occurrences of 2) = Number of favorable outcomes / Total number of possible outcomes
P(of exactly 4 occurrences of 2) = 90 / 46656 ≈ 0.00193 (rounded to 5 decimal places).
Therefore, the probability of rolling exactly 4 occurrences of the number 2 when rolling a die 6 times is approximately 0.00193.