Answer:
a) 26
b) 0.3076
Step-by-step explanation:
We are given the following in the question:
M: Math
P: Physics
S: Statistic
n(Math) = 24
n(Physics) = 29
[tex]n(P\cap S) = 10[/tex]
[tex]n(M\cap P) = 13[/tex]
[tex]n(M\cap S) = 11[/tex]
[tex]n(M\cap P\cap S) = 8[/tex]
[tex]n(M'\cap P'\cap S') = 7[/tex]
a) Number of students who take statistics
Formula:
[tex]n(P\cup M\cup S)=\\n(P)+n(M)+n(S)-n(P\cap M)-n(P\cap S)-n(M\cap S)+n(P\cap M\cap S)[/tex]
Putting the values, we get,
[tex]n(P\cup M\cup S) = n - n(P'\cap M'\cap S') = 60-7 = 53[/tex]
[tex]53 = 24 + 29 + n(S) -10-13-11+8\\n(S) = 53-24-29+10+13+11-8\\n(S) = 26[/tex]
Thus, 26 students took statistic.
b) probability that a student selected at random takes all three, given he takes statistics
[tex]p(\text{All three subject}|\text{Statistic})\\\\=\displaystyle\frac{n(P\cap M\cap S)\cap n(S)}{n(S)}\\\\= \frac{8}{26} = 0.3076[/tex]