Respuesta :
Answer:
1/4669920
Step-by-step explanation:
First if all, when repetition is mot allowed, we use Permutation to get the number of possible outcomes as follows;
P(n, r) = n!/(n - r)!
Now, the order is important since the first, second, third and fourth prizes are different.
Thus, we need to define the Permutation.
We need to select 4 winners of the 48 people.
Thus, n = 48 and r = 4
Thus;
P(48,4) = 48!/(48 - 4)!
P(48,4) = 4669920
Now, probability is the number of favourable outcomes divided by number of possible outcomes
Thus,
Probability = 1/4669920
The probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively is ( 1 / (2^(48) - 1) )^4
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
For this case, we're given that:
- first, second, third and fourth prizes are there.
- 48 people entered the contest.
- winning more than one prize is allowed
We assume that all 48 people are equally probable to get any of those prizes.
First prize can be won by 48 people, and so as second third and fourth.
That makes all of the prizes' winners independent of each other.
Think of it as prizes themselves touching people, and more than one prize can touch one specific people.
So, we have:
P(Bo getting first, Colleen getting second, Jeff getting third and Rohini getting fourth prize) = P(first prize touching Bo, second prize touching Colleen, third prize touching Jeff, fourth prize touching Rohini).
Assuming that Bo getting first prize means no one else gets it, and so for 2nd, 3rd and fourth prize, its like we restrict each of first prize, second prize, 3rd and 4th prize to touch only 1 person out of those 48 choices.
Thus, for first prize and so as for any prize:
P(First prize choosing only 1 person) = 1/ number of ways it can choose any number of people out of those 48 people.
Number of ways first prize can choose any number of people (from 1 to 48) out of those 48 people is:
[tex]\sum_{n=1}^{48}(\: ^{48}C_n} = 2^{48} - 1[/tex]
Thus, by using the chain rule, we get:
P(P(Bo getting first, Colleen getting second, Jeff getting third and Rohini getting fourth prize) = [tex]\dfrac{1}{2^{48} -1} \times \dfrac{1}{2^{48} -1} \times\dfrac{1}{2^{48} -1} \times\dfrac{1}{2^{48} -1} = \left (\dfrac{1}{2^{48} -1} \right )^4[/tex]
Thus, the probability that Bo, Colleen, Jeff, and Rohini win the first, second, third, and fourth prizes, respectively is ( 1 / (2^(48) - 1) )^4
Learn more about probability here:
brainly.com/question/1210781
