Question

Each of 100 students in the Allen School can only take 1 CSE class each, between the four classes CSE 311, CSE 312, CSE 331, and CSE 332. Each student (independently of others) takes CSE 311 with probability 0.3, CSE 312 with probability 0.4, CSE 331 with probability 0.1, and CSE 332 with probability 0.2. What is the probability that exactly 31 sign up for CSE 311, 39 sign up for CSE 312, 7 sign up for CSE 331, and 23 sign up for CSE 332

Answer 1

Answer:

$P(a=31,b=39,c=7,d=23) = 0.000668$

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, $P_a$ = 0.3

Probability of taking CSE 312, $P_b$ = 0.4

Probability of taking CSE 331, $P_c$ = 0.1

Probability of taking CSE 332, $P_d$ = 0.2

$P(a,b,c,d) = \frac{n!}{a! b! c! d!} p_a^{a} p_b^{b} p_c^{c} p_d^{d} \\P(a=31,b=39,c=7,d=23) = \frac{100!}{31! 39! 7! 23!} * 0.3^{31} * 0.4^{39} * 0.1^{7} 0.2^{23}\\P(a=31,b=39,c=7,d=23) = \frac{4.58*10^{111}}{2.13*10^{56}* 5040 }* (1.57*10^{-55})\\P(a=31,b=39,c=7,d=23) = 0.000668$