Please provide the solution to the below problems:
In a certain state lottery, a player selects four different numbers from 1 through 30. Find the probability that a single choice of four numbers wins the lottery, assuming the order of the numbers is not important.
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A club consisting of 20 members wishes to form a four-member committee. In how many ways can this be done?
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A jar contains 4 red balls and 6 green balls. A second jar contains 5 red balls and 8 black balls. If we select one ball from each jar, what is the probability of obtaining 1 red ball and 1 black ball?
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What is the probability of flipping a coin five times and obtaining exactly three heads?
You are choosing 4 from 30 numbers
= C(30,4) = 30!/(4!26!) = 27405
one of those will be correct
prob(correct #) = 1/27405
2. C(20,4) = ...
3. You don't say which jar we choose from first
but the first jar does not contain a black ball, so the order must be jar1, then jar 2
prob(red,black) = (4/10)(8/13) = 16/65
4. Prob(3 heads out of 5)
= C(5,3) (1/2)^3 (1/2)^2
= 10(1/32) = 5/16
To solve the problems, we need to use basic principles of combinatorics and probability.
1. In the first problem, the player selects four different numbers from 1 through 30. Since the order of the numbers selected does not matter, we need to calculate the probability of selecting the winning numbers, regardless of their order.
To calculate the probability, we need to determine the number of favorable outcomes (winning combinations) and the total number of possible outcomes (all combinations).
The number of favorable outcomes is 1, as only one set of numbers can win the lottery.
To calculate the total number of possible outcomes, we need to use combination formula. The total number of possible combinations can be found by calculating "30 choose 4" or C(30, 4). This can be calculated as:
C(30, 4) = 30! / (4! * (30-4)!) = 27,405
Therefore, the probability of winning the lottery is 1/27,405.
2. In the second problem, the club consisting of 20 members wants to form a four-member committee. The order of selection is not important.
Here, we need to calculate the total number of possible combinations. This can be found using the combination formula, where the total number of possible combinations is "20 choose 4" or C(20, 4). This can be calculated as:
C(20, 4) = 20! / (4! * (20-4)!) = 4,845
Therefore, there are 4,845 possible ways to form a four-member committee.
3. In the third problem, there are two jars containing red and green balls as well as red and black balls, respectively. We select one ball from each jar and need to calculate the probability of obtaining 1 red ball and 1 black ball.
To calculate the probability, we need to determine the number of favorable outcomes and the total number of possible outcomes.
The number of favorable outcomes is the number of ways of selecting 1 red ball from the first jar (4 options) multiplied by the number of ways of selecting 1 black ball from the second jar (8 options).
The total number of possible outcomes is the number of ways of selecting 1 ball from each jar, which is the product of the number of balls in each jar (6 * 13).
Therefore, the probability of obtaining 1 red ball and 1 black ball is (4/6) * (8/13) = 32/78.
4. In the fourth problem, we need to calculate the probability of flipping a coin five times and obtaining exactly three heads.
To calculate the probability, we need to determine the number of favorable outcomes and the total number of possible outcomes.
The number of favorable outcomes is the number of ways of getting exactly three heads when flipping a coin five times, which can be calculated using the binomial probability formula.
The total number of possible outcomes is the total number of ways of flipping a coin five times, which is 2^5 (since each flip has two possible outcomes, heads or tails).
Therefore, the probability of obtaining exactly three heads is (5 choose 3) * (1/2)^3 * (1/2)^2 = 10/32, or 5/16.