The probability of drawing a student from the acers on the first try is 78/400 or 0.195.
Since the computer program allows for the same student to be drawn twice, the total number of students remains 400 for the second draw. However, since one student has already been drawn from the acers, there are now only 77 students left in that team.
Therefore, the probability of drawing a student from the acers on the second try, given that a student has already been drawn from that team on the first try, is 77/400 or 0.1925.
Stillwater Junior High divides students into teams taught by a group of teachers. The table shows the number of students in each time.
Team: Number of students:
Acers 78
Blazers 80
Outbacks 83
Quasars 77
Voyagers 82
The total number of students is 400.
The principal uses a computer to randomly select the name of a student from all the students in the school. With the computer program, it is possible to draw the name of the same student twice. If the principal selects the name of a student from the acers on the first try, what is the probability she will draw the name of a student from the acers on the second try?
5 answers
are you sure
Yes, the probability of drawing a student from the Acers on the second try, given that a student has already been drawn from that team on the first try, is 77/400 or 0.1925.
On the first draw, there are 78 students in the Acers team out of a total of 400 students, so the probability of selecting a student from the Acers on the first draw is 78/400 or 0.195.
On the second draw, since the computer program allows for the same student to be drawn twice, there are still 400 students in the pool, but only 77 students left in the Acers team since one student has already been drawn. Therefore, the probability of selecting a student from the Acers on the second draw, given that a student has already been drawn from that team on the first draw, is 77/400 or 0.1925.
On the first draw, there are 78 students in the Acers team out of a total of 400 students, so the probability of selecting a student from the Acers on the first draw is 78/400 or 0.195.
On the second draw, since the computer program allows for the same student to be drawn twice, there are still 400 students in the pool, but only 77 students left in the Acers team since one student has already been drawn. Therefore, the probability of selecting a student from the Acers on the second draw, given that a student has already been drawn from that team on the first draw, is 77/400 or 0.1925.
among the options a. 1/78 b. 38/399 c. 19/100 d. 39/200 which is correct
The correct option is (b) 38/399.
The probability of drawing a student from the Acers on the first try is 78/400 or 0.195.
Since one student has already been drawn from the Acers, there are now only 77 students left in that team. Therefore, on the second draw, the probability of drawing the same student again is 1/77 (since there is only one student who has already been drawn and there are 77 students left in the team).
However, we need to find the probability of drawing any student from the Acers on the second try given that a student has already been drawn from that team on the first try. This can be calculated using the conditional probability formula:
P(A | B) = P(A and B) / P(B)
where P(A | B) is the probability of A given B, P(A and B) is the probability of both A and B occurring, and P(B) is the probability of B occurring.
In this case, A represents the event of drawing a student from the Acers on the second try and B represents the event of drawing a student from the Acers on the first try. Using the formula, we get:
P(A | B) = (1/77) / (78/400)
= (1/77) x (400/78)
= 400/6006
= 38/399
Therefore, the correct option is (b) 38/399.
The probability of drawing a student from the Acers on the first try is 78/400 or 0.195.
Since one student has already been drawn from the Acers, there are now only 77 students left in that team. Therefore, on the second draw, the probability of drawing the same student again is 1/77 (since there is only one student who has already been drawn and there are 77 students left in the team).
However, we need to find the probability of drawing any student from the Acers on the second try given that a student has already been drawn from that team on the first try. This can be calculated using the conditional probability formula:
P(A | B) = P(A and B) / P(B)
where P(A | B) is the probability of A given B, P(A and B) is the probability of both A and B occurring, and P(B) is the probability of B occurring.
In this case, A represents the event of drawing a student from the Acers on the second try and B represents the event of drawing a student from the Acers on the first try. Using the formula, we get:
P(A | B) = (1/77) / (78/400)
= (1/77) x (400/78)
= 400/6006
= 38/399
Therefore, the correct option is (b) 38/399.
5 answers