In the game of roulette a ball rolls into one of 37 slots. All the slots are evenly likely. Eighteen of the slots are red, eighteen are white and 1 is green. the ball is rolled 10 times. Calculate the probability of exactly 6 reds? the answer is 0.l94
To calculate the probability of exactly 6 reds in 10 rolls of the roulette ball, you can use the binomial probability formula:
P(x) = (nCx) * (p^x) * (q^(n-x))
Where:
P(x) is the probability of getting exactly x successes (in this case, exactly 6 reds),
n is the total number of trials (in this case, 10 rolls),
x is the number of successful outcomes you're interested in (in this case, 6 reds),
p is the probability of a single success (in this case, the probability of getting a red slot in one roll), and
q is the probability of a single failure (in this case, the probability of getting a non-red slot in one roll).
In this scenario, there are 18 red slots out of a total of 37 slots, so the probability of getting a red slot in one roll is calculated as:
p = 18/37
Similarly, the probability of getting a non-red (white or green) slot is calculated as:
q = 1 - p = 1 - (18/37)
Plugging in the values into the formula, we get:
P(6) = (10C6) * ((18/37)^6) * ((1 - (18/37))^(10-6))
Using the formula for combinations (nCx) = n! / (x!(n-x)!), we can calculate (10C6) as:
(10C6) = 10! / (6!(10-6)!) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 210
Substituting all the values into the formula:
P(6) = 210 * ((18/37)^6) * ((1 - (18/37))^(10-6))
Calculating this expression gives us approximately 0.194, which is rounded to 0.194 or 0.19.
Therefore, the probability of exactly 6 reds in 10 rolls of the roulette ball is 0.194 or approximately 0.19.
prob(red) = 18/37
prob(not red) = 19/37
prob 6 reds out of 10
= C(10,6) (18/37)^6 (19/37)^4
= using calculator
= 210(.00092179)
= appr .194