Shankar is teaching 2 courses, let us call them 435 and FYS. 435 has 3 sophomores, 8 juniors and 13 seniors; FYS has 5 sophomores, 7 juniors and 6 seniors. Suppose he wants to understand how his teaching is for the students. He picks one student from each class at random. Find the probability that both students are of the same type (i. e. both sophomores, or both juniors etc). Simplify your answer so it looks like a fraction without any factorials.

The probability that both students are of the same type is .

Step-by-step explanation:

The students in 435 are: {3 sophomores, 8 juniors and 13 seniors}

Number of students in 435 = 3 + 8 + 13 = 24

The students in FYS are: {5 sophomores, 7 juniors and 6 seniors}.

Number of students in FYS = 5 + 7 + 6 = 18

The teacher picks 1 student from each class.

The probability that both students are of the same type is:

P (Same type students) = P (Both are Sophomores) + P (Both are Juniors)

                                                     + P (Both are Seniors)

= P (Sophomore ∩ Course 435) × P (Sophomore ∩ Course FYS)

         + P (Junior ∩ Course 435) × P (Junior ∩ Course FYS)

               + P (Senior ∩ Course 435) × P (Senior ∩ Course FYS)

Thus, the probability that both students are of the same type is .

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

answer4 to the second power is 16

and 4 to the third power is 64,