Question

A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 comma 7 comma 8 comma 9 comma 10 comma 11 comma 12 comma 13 comma 14 comma 15 endsets={4,5,6,7,8,9,10,11,12,13,14,15}. let event upper e equals startset 7 comma 8 comma 9 comma 10 comma 11 comma 12 comma 13 comma 14 comma 15 endsete={7,8,9,10,11,12,13,14,15}. assume each outcome is equally likely. list the outcomes in upper e superscript cec. find upper p left parenthesis upper e superscript x right parenthesis spec.

Answer 1

The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in $E^{c}$. Find $P(E^{c})$.

Solution:

Part 1:

$E^{c}$ means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

$E^{c}=S-E \\ \\ E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\ \\ E^{c}=(4,5,6)$

Thus the set compliment of E will contain the elements {4,5,6}.So

$E^{c}$ = {4,5,6}

Part 2)

$P(E^{c})$ means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

$P(E^{c})$ = (Number of Elements in $E^{c}$)/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set $E^{c}$ = $n(E^{c})$=3

So, 

$P(E^{c})= \frac{n(E^{c}) }{n(S)} \\ \\ P(E^{c})= \frac{3}{12} \\ \\ P(E^{c})= \frac{1}{4}$