Respuesta :
Answer:
a) 16.7% probability that E will speak first.
b) 3.33% probability that C will speak fifth and B will speak last.
c) 0.14% probability that the students will speak in the following order: DECABF.
d) 33.33% probability that A or B will speak first.
Step-by-step explanation:
The first speaker can be any of them. The second can be any of them bar the first. It continues until the last one. So the total number of outcomes is 6! = 720.
(a) E will speak first
The first one to speak is E.
The other five can be selected in any order. So there are 5! = 120 total outcomes in which E speaks first.
There is a 120/720 = 0.167 probability that E will speak first.
(b) that C will speak fifth and B will speak last
The fifth can only be C and the sixth can only be B.
The first four can be selected in any order. So there are 4! = 24 outcomes in which C speaks fifth and B last.
There is a 24/720 = 0.0333 probability that C will speak fifth and B will speak last.
(c) that the students will speak in the following order: DECABF
This is the only possible outcome in this order.
So there is a 1/720 = 0.0014 = 0.14% probability that the students will speak in the following order: DECABF.
(d) that A or B will speak first.
For A first, the following five can be in any order. So there are 5! = 120 outcomes in which A speaks first. The same logic and result for B.
So there are 240 outcomes in which A or B speaks first.
There is a 240/720 = 0.3333 = 33.33% probability that A or B will speak first.
The solution tp the different probabilities based on the given speakers are;
a) P(E will speak first) = 0.167
b) P(that C will speak fifth and B will speak last) = 0.033
c) P(that the students will speak in the following order: DECABF) = 0.0014
d) P(that A or B will speak first) = 0.333
There are 6 students namely A, B, C, D, E and F to give speeches
Since the order of speaking is random, then total possible ways they can be arranged is; 6! = 720 ways.
(a) Probability that E will speak first.
If E speaks first, then the five other speakers can be arranged in;
5! ways = 5 × 4 × 3 × 2 × 1 = 120 ways
Thus;
P(E will speak first) = 120/720
P(E will speak first) = 0.167
(b) Probability that C will speak fifth and B will speak last
This means that the fifth and last positions are fixed.
Thus, the number of ways that the 4 other positions can be arranged is;
4! = 24 ways
Thus;
P(that C will speak fifth and B will speak last) = 24/720
P(that C will speak fifth and B will speak last) = 0.033
(c) Probability that the students will speak in the following order: DECABF
This means that there is only way way in which they can all speak.
Thus;
P(that the students will speak in the following order: DECABF) = 1/720
P(that the students will speak in the following order: DECABF) = 0.0014
(d) Probability that A or B will speak first.
If A speaks first, then the five other speakers can be arranged in;
5! ways = 5 × 4 × 3 × 2 × 1 = 120 ways
Also, if B speaks first, then the five other speakers can be arranged in;
5! ways = 5 × 4 × 3 × 2 × 1 = 120 ways
Total ways in which A or B can speak first = 120 + 120 = 240 ways
Thus;
P(that A or B will speak first) = 240/720
P(that A or B will speak first) = 0.333
Read more at; https://brainly.com/question/16315950
