Answer:
[tex]P(a=31,b=39,c=7,d=23) = 0.000668[/tex]
Step-by-step explanation:
Sample space, n = 100
Let the number of students signed up for CSE 311 = a
Let the number of students signed up for CSE 312 = b
Let the number of students signed up for CSE 331 = c
Let the number of students signed up for CSE 332 = d
Probability of taking CSE 311, [tex]P_a[/tex] = 0.3
Probability of taking CSE 312, [tex]P_b[/tex] = 0.4
Probability of taking CSE 331, [tex]P_c[/tex] = 0.1
Probability of taking CSE 332, [tex]P_d[/tex] = 0.2
[tex]P(a,b,c,d) = \frac{n!}{a! b! c! d!} p_a^{a} p_b^{b} p_c^{c} p_d^{d} \\P(a=31,b=39,c=7,d=23) = \frac{100!}{31! 39! 7! 23!} * 0.3^{31} * 0.4^{39} * 0.1^{7} 0.2^{23}\\P(a=31,b=39,c=7,d=23) = \frac{4.58*10^{111}}{2.13*10^{56}* 5040 }* (1.57*10^{-55})\\P(a=31,b=39,c=7,d=23) = 0.000668[/tex]
