Answer: [tex]\frac{1}{15}[/tex]
Explanation:
Total number of students are six
Number of students who forgot their lunch is 2
We need to select the students who forgot their lunch so, required probability would be[tex]\frac{^2C_2\cdot ^4C_0}{^6C_6}[/tex]
here [tex]^2C_2[/tex] because we are selecting 2 students among 6 who forgot their lunch students who forgot their lunch is 2
[tex]^4C_0[/tex] because number of students who did not forgot their lunch is 4 and we will not select the student who bring lunch hence 0
[tex]^6C_2[/tex] we are selecting two students out of total number of students
Hence, after simplification of the below expression
[tex]\frac{^2C_2\cdot ^4C_0}{^6C_6}[/tex][tex]=\frac{2!4!}{6!}=\frac{2}{30}= \frac{1}{15}[/tex]
The probability that both of the considered students forgot their lunch for this case is 1/15
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.
Out of 6 students, the number of ways teacher can choose 2 students is:
[tex]^6C_2 = \dfrac{6 \times 5}{2 \times 1} = 15\: \rm ways[/tex]
The number of ways teacher can select those 2 students who forgot their lunch = 1 (only 1 such way exist).
Thus, if we take:
E = event that the 2 students chosen have forgot their lunch, then:
n(E) = 1
n(S) = 15
and p(E) = 1/15
Thus, the probability that both of the considered students forgot their lunch for this case is 1/15
Learn more about probability here:
brainly.com/question/1210781
