An ordinary die is thrown seven times.find the probability of obtaining exactly three sixes
prob(six) = 1/6
prob(not six) = 5/6
prob(as stated) = C(7,3) (1/6)^3 (5/6)^4
= 35(1/216)(625/1296) = ....
Well, the probability of obtaining a six on a single throw of a fair die is 1/6. Therefore, in order to obtain exactly three sixes in seven throws, we can use the binomial probability formula.
Using the formula, we can calculate the probability using the following equation:
P(X = k) = (n! / (k! * (n - k)!)) * (p^k) * ((1-p)^(n-k))
where:
P(X = k) is the probability of getting exactly k successes (in this case, exactly three sixes),
n is the total number of trials (seven throws of the die),
k is the number of successful outcomes (three sixes),
p is the probability of a single successful outcome (1/6).
Plugging in the values, we have:
P(X = 3) = (7! / (3! * (7-3)!)) * ((1/6)^3) * ((1 - 1/6)^(7-3))
Simplifying the equation, we get:
P(X = 3) = (35 / 6,160) * (1/216) * (125/216)
Calculating further, we have:
P(X = 3) ≈ 0.0143
So, the probability of obtaining exactly three sixes in seven throws is approximately 0.0143 or 1.43%.
But honestly, when it comes to dice, you never really know what'll happen. Maybe the dice will take a liking to the number six and roll it more often just to mess with you. It's a dicey situation, you know? *ba dum tss*
To find the probability of obtaining exactly three sixes when throwing an ordinary die seven times, we need to calculate the total number of favorable outcomes and divide it by the total number of possible outcomes.
1. Firstly, let's find the total number of possible outcomes for throwing a die seven times. Since each throw has 6 possible outcomes (numbers 1 to 6), the total number of possible outcomes is 6^7 = 279,936.
2. Now, let's determine the number of favorable outcomes, which means getting exactly three sixes. We can break this down into three parts: the three positions that have a six and the four positions that don't.
a. The three positions that have a six can be chosen in C(7, 3) = 35 ways. (This is the number of combinations of 7 items taken 3 at a time.)
b. For each position with a six, there is only one favorable outcome (getting a six).
c. The remaining four positions can have any number from 1 to 5 (since we already accounted for the three sixes). Thus, there are 5^4 = 625 favorable outcomes for these positions.
3. The total number of favorable outcomes is the product of the number of favorable outcomes for each part: 35 * 1 * 625 = 21,875.
4. Finally, we divide the total number of favorable outcomes (21,875) by the total number of possible outcomes (279,936) to find the probability:
P(exactly 3 sixes) = 21,875 / 279,936 ≈ 0.0781
Therefore, the probability of obtaining exactly three sixes when throwing an ordinary die seven times is approximately 0.0781 or 7.81%.
To find the probability of obtaining exactly three sixes when throwing an ordinary die seven times, we need to calculate the probability of getting a six on exactly three throws.
Step 1: Determine the probability of getting a six on a single throw.
Since there are six outcomes on a die (numbers 1-6) and only one of them is a six, the probability of getting a six on a single throw is 1/6.
Step 2: Determine the probability of getting a non-six on a single throw.
Since there are six outcomes on a die (numbers 1-6) and five of them are non-sixes, the probability of getting a non-six on a single throw is 5/6.
Step 3: Determine the number of ways to get exactly three sixes in seven throws.
To calculate the number of ways to get exactly three sixes in seven throws, we use the formula for combinations. This can be written as:
C(n, k) = n! / (k!(n - k)!)
where n is the total number of throws and k is the number of successful outcomes (in this case, getting a six). In our case, it would be:
C(7, 3) = 7! / (3!(7 - 3)!) = 7! / (3!4!)
Step 4: Calculate the probability of getting exactly three sixes.
To calculate the probability of obtaining exactly three sixes, we multiply the probability of getting a six on a single throw (1/6) by the probability of getting a non-six on a single throw (5/6) for the remaining non-six outcomes. Finally, we multiply this by the number of ways to get exactly three sixes in seven throws.
Probability = (1/6)^3 * (5/6)^(7-3) * C(7, 3)
Plugging in the values, we get:
Probability = (1/6)^3 * (5/6)^4 * 35
Probability = 0.0046
Therefore, the probability of obtaining exactly three sixes when throwing an ordinary die seven times is approximately 0.0046 or 0.46%.