The author purchased a slot machine configured so that there is a 1/2000 probability of winning the jackpot on any individual trial. Although no one would seriously consider tricking the author, suppose that a guest claims that she played the slot machine 5 times and hit the jackpot twice.
Find the probability of at least 2 jackpots in 5 trials
C(5,2) = 5!/[3!(2!)] = 5*4 /2 = 10
P(2 in 5 trials) = 10 (1/2000)^2 (1-1/2000)^3
(1-1/2000)^3 is about 1
so
10/(2 *10*3)^2 = (10/4)* 10^-6
= 2.5 * 10^-6 = 0.0000025
Thank you...
To find the probability of getting at least 2 jackpots in 5 trials, we need to find the probability of getting exactly 2 jackpots, exactly 3 jackpots, 4 jackpots, and 5 jackpots, and then add them all together.
The probability of getting exactly 2 jackpots in 5 trials can be calculated using the binomial probability formula:
P(X=k) = (nCk) * (p^k) * (1-p)^(n-k)
Where:
P(X=k) is the probability of getting exactly k jackpots
n is the number of trials (5 in this case)
k is the number of jackpots (2 in this case)
p is the probability of winning the jackpot on any individual trial (1/2000 in this case)
Using this formula, we can calculate the probability of getting exactly 2 jackpots:
P(X=2) = (5C2) * ((1/2000)^2) * (1-(1/2000))^(5-2)
= (10) * ((1/2000)^2) * ((1999/2000)^(3))
= 10 * (1/4000000) * (7960196001/8000000000)
= 0.019900495123
Similarly, we can calculate the probabilities of getting exactly 3, 4, and 5 jackpots using the same formula:
P(X=3) = (5C3) * ((1/2000)^3) * (1-(1/2000))^(5-3)
= (10) * ((1/2000)^3) * ((1999/2000)^(2))
= 10 * (1/8000000000) * (1593601601/1600000000)
= 0.000994502488
P(X=4) = (5C4) * ((1/2000)^4) * (1-(1/2000))^(5-4)
= (5) * ((1/2000)^4) * ((1999/2000)^(1))
= 5 * (1/16000000000) * (998000001/1000000000)
= 0.000031124268
P(X=5) = (5C5) * ((1/2000)^5) * (1-(1/2000))^(5-5)
= (1) * ((1/2000)^5) * ((1999/2000)^(0))
= 1 * (1/32000000000000)
= 0.00000000003125
Finally, we can add up these probabilities to get the probability of at least 2 jackpots in 5 trials:
P(at least 2 jackpots) = P(X=2) + P(X=3) + P(X=4) + P(X=5)
= 0.019900495123 + 0.000994502488 + 0.000031124268 + 0.00000000003125
≈ 0.02092612191
Therefore, the probability of getting at least 2 jackpots in 5 trials is approximately 0.0209 (or 2.09%).
To find the probability of at least 2 jackpots in 5 trials, we can use the binomial probability formula. The binomial probability formula is given by:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting x successes in n trials,
C(n, x) is the number of combinations of n items taken x at a time,
p is the probability of success on a single trial, and
(1-p) is the probability of failure on a single trial.
In this case, the probability of winning the jackpot on any individual trial is 1/2000. So, the probability of not winning the jackpot on any individual trial is 1 - 1/2000 = 1999/2000.
Now, we need to calculate the probability of getting at least 2 jackpots in 5 trials. This can be done by summing the probabilities of getting exactly 2, 3, 4, and 5 jackpots.
P(at least 2 jackpots in 5 trials) = P(2 jackpots) + P(3 jackpots) + P(4 jackpots) + P(5 jackpots)
P(2 jackpots) = C(5, 2) * (1/2000)^2 * (1999/2000)^(3)
P(3 jackpots) = C(5, 3) * (1/2000)^3 * (1999/2000)^(2)
P(4 jackpots) = C(5, 4) * (1/2000)^4 * (1999/2000)^(1)
P(5 jackpots) = C(5, 5) * (1/2000)^5 * (1999/2000)^(0)
Using the formula for combinations (C(n, x) = n! / (x! * (n-x)!)), we can calculate the individual probabilities:
P(2 jackpots) = (5! / (2! * (5-2)!)) * (1/2000)^2 * (1999/2000)^3
P(3 jackpots) = (5! / (3! * (5-3)!)) * (1/2000)^3 * (1999/2000)^2
P(4 jackpots) = (5! / (4! * (5-4)!)) * (1/2000)^4 * (1999/2000)^1
P(5 jackpots) = (5! / (5! * (5-5)!)) * (1/2000)^5 * (1999/2000)^0
Now, we can substitute the values and calculate the probabilities:
P(2 jackpots) = (5! / (2! * 3!)) * (1/2000)^2 * (1999/2000)^3
P(3 jackpots) = (5! / (3! * 2!)) * (1/2000)^3 * (1999/2000)^2
P(4 jackpots) = (5! / (4! * 1!)) * (1/2000)^4 * (1999/2000)^1
P(5 jackpots) = (5! / (5! * 0!)) * (1/2000)^5 * (1999/2000)^0
After calculating these probabilities individually, we can sum them to get the final result:
P(at least 2 jackpots in 5 trials) = P(2 jackpots) + P(3 jackpots) + P(4 jackpots) + P(5 jackpots)