A fair six sided dice has the faces numbered from 1 to 6. The dice is thrown 4 time. Calculate the probability of exactly 2 fives. the answer is 25/216
prob(5) = 1/6
prob(not 5) = 5/6
you want 5 twice, not5 2 twice
prob = C(4,2) (1/6)^2 (5/6)^2
= 6(1/36)(25/36) = 25/216
Well, isn't that a roll of the dice! To calculate the probability of exactly 2 fives, we need to consider the number of possible outcomes and the number of favorable outcomes.
The number of possible outcomes when throwing a six-sided dice 4 times is 6^4, which is equal to 1296.
Now, let's break it down. To get exactly 2 fives, we can consider the following scenarios:
1) FFFD: This means getting a five three times (FFF) and any other number (D) once.
2) FFD: Getting a five two times (FF) and any other number once (D).
3) FDF: Getting a five once (F), any other number once (D), and a five again (F).
4) DFF: Getting any other number once (D), a five once (F), and a five again (F).
So, we have four favorable outcomes.
Therefore, the probability of exactly 2 fives is 4/1296, which can be simplified to 1/324.
Hmmm, that's not quite the same as your answer of 25/216. It seems I might have rolled my calculations wrong. Thanks for pointing that out, and please forgive my mathematical clowning around!
To calculate the probability of exactly 2 fives when throwing a fair six-sided dice 4 times, we can use the concept of binomial distribution.
The binomial distribution is given by the 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 binomial coefficient, calculated as n! / (k!(n-k)!)
- p is the probability of getting a success in a single trial
- n is the total number of trials
In this case, we want exactly 2 fives, so k=2, and we have 4 trials (n=4). The probability of getting a five (success) on a fair six-sided dice is 1/6 (p=1/6).
Plugging in these values into the formula:
P(X=2) = (4 choose 2) * (1/6)^2 * (5/6)^(4-2)
Calculating the binomial coefficient:
(4 choose 2) = 4! / (2!(4-2)!)
= 4! / (2! * 2!)
= 4 * 3 / (2 * 1)
= 6
Now, let's calculate the probability:
P(X=2) = 6 * (1/6)^2 * (5/6)^(4-2)
= 6 * (1/36) * (25/36)
= 150/1296
= 25/216
Therefore, the probability of exactly 2 fives when throwing a fair six-sided dice 4 times is 25/216.
To calculate the probability of exactly 2 fives when a fair six-sided dice is thrown 4 times, we need to determine the number of favorable outcomes (i.e., the number of ways to get exactly 2 fives) and the total number of possible outcomes (i.e., the number of ways the dice can land on any number from 1 to 6, four times in a row).
Step 1: Determine the number of favorable outcomes (ways to get exactly 2 fives).
To get exactly 2 fives, we can consider the two fives as a special event, and the remaining two throws as any number from 1 to 6. The two non-five throws can consist of any number from 1 to 6 in each of the two throws, except for 5.
The number of ways to choose the two non-five throws is given by the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the total number of options (6 in this case) and r is the number of selections (2 in this case).
Therefore, the number of ways to choose the two non-five throws is C(6, 2) = (6! / (2! * (6-2)!)) = 15.
For each of these 15 combinations, we have only one possibility for the two fives. So, the number of favorable outcomes is 15.
Step 2: Determine the total number of possible outcomes.
When a six-sided dice is thrown 4 times, each of the four throws can result in any number from 1 to 6. So, for each throw, we have 6 possibilities.
Thus, the total number of possible outcomes is 6^4 = 1296.
Step 3: Calculate the probability.
The probability of an event occurring is given by the formula: Probability = (Number of favorable outcomes) / (Total number of possible outcomes).
So, the probability of getting exactly 2 fives when a fair six-sided dice is thrown 4 times is:
Probability = 15/1296 = 1/86.