Question

Let x be the number of courses for which a randomly selected student at a certain university is registered. The probability distribution of x appears in the table shown below:

x 1 2 3 4 5 6 7

p(x) .05 .03 .09 .26 .37 .16 .04

(a) What is P(x = 4)? P(x = 4) =

(b) What is P(x 4)? P(x 4) =

(c) What is the probability that the selected student is taking at most five courses? P(at most 5 courses) =

(d) What is the probability that the selected student is taking at least five courses?

more than five courses?

P(at least 5 courses) =

P(more than 5 courses) =

(e) Calculate P(3 x 6) and P(3 < x < 6). P(3 x 6) = P(3 < x < 6) =

Answer 1

Here, X is the no. of courses for which a randomly selected student at a certain university is registered.

The probability distribution of X is being given in the table :-

X	1	2	3	4	5	6	7
p(X)	0.05	0.03	0.09	0.26	0.37	0.16	0.04

a) Here, we have been given a problem to find P[X=4].

Now here, the value of p(X) corresponding to X=4 is 0.26.Hence, the value of P[X=4], i.e. the probability that a student has applied for four courses in that certain university is 0.26.

Therefore, P[X=4]=0.26

b) Here we have been given a problem to find P[X4]=P[X<4]

Now,in order to find out the probability that a student applies for less than four coursesin that certain university, we have to add up the p(X) values for X<4.

Here, P[X=1]=0.05

P[X=2]=0.03

P[X=3]=0.09

Hence the probability that a student has applied for less than four courses in the university is given by

P[X<4]= P[X=1] + P[X=2] + P[X=3] = 0.05 + 0.03 + 0.09 = 0.17 Therefore, P[X<4]=P[X4]=0.17 c) Here we have been given a problem to calculate the probability that a selected student is taking at most five courses. Now, in order to find out the probability that a student takes up atmost 5 courses, we have to take into account that the student might take up lesser than five courses too. Hence, we have to calculate the measure P[X<=5]. Now, P[X=1]=0.05, P[X=2]=0.03, P[X=3]=0.09, P[X=4]=0.26, P[X=5]=0.37 Hence P[at most 5 courses] = P[X<=5] = P[X=1] + P[X=2] + P[X=3] + P[X=4] + P[X=5] = 0.05 + 0.03 + 0.09 + 0.26 + 0.37 = 0.80

Hence P[at most 5 courses] = 0.80

d) Here, we have been given two problems i) To find out the probability that a student takes up at least 5 courses. ii) To find out the probability that a student takes up more than 5 courses.

i) Now, in order to find out the probability that a student takes up at least 5 courses, we have to calculate the measure P[X>=5]. Now, P[X=5]=0.37, P[X=6]=0.16 , P[X=7]=0.04 Hence P[at least 5 courses] = P[X>=5] = P[X=5] + P[X=6] + P[X=7] = 0.37 + 0.16 + 0.04 = 0.57 Therefore P[at least 5 courses] = 0.57

ii) Here, in order to calculate the probability that a student takes up more than 6 courses, we have to calculate the measure P[X>5]. Hence, P[more than 5 courses] = P[X>5] = P[X=6] + P[X=7] = 0.16 +0.04 = 0.20 Therefore, P[more than 5 courses] = 0.20

e) Now, here we have been given a problem to calculate P[3 X 6] = P[3<X<6] i.e. to find out the probability that a student takes up the no. of courses between three and six. Hence, in such a scenario, we can only accept cases where a student takes up four courses or five courses. Here P[X=4] = 0.26 and P[X=5] = 0.37 Hence P[ 3 X 6 ] = P[3<X<6] = P[X=4] + P[X=5] = 0.26 + 0.37 =0.63 Therefore P[ 3 X 6 ] = P[3<X<6] = 0.63