Question

The classrooms at the Redman Academy consist of cramped tables that seat 3 people. The probability of being-left handed is 26%, and left-handed test takers need to sit in the far-left seat, otherwise, they constantly bump elbows with the person next to them causing complaints.

A) What is the probability that no one in a group of 3 students is left-handed?

B) What is the probability that only 1 person in the group of 3 students is left-handed?

C) What is the probability that no one at a table will complain during a test if I randomly seat three students?

D) I have 9 students in my class, 2 of which are left-handed. If I have exactly 3 tables and randomly seat the students, what is the probability of no complaints?

Answer 1

Let L = Event that the person is left handed and R = Event that the person is right handed

Given Pr(L) = 0.26

By definition of complementary probability,

Pr( L') = Probability that the person is not left handed

= 1 - P(L)

= 1 - 0.26

= 0.74

A) The probability that no one in a group of 3 students is left-handed

(R,R,R)

= 0.74 x 0.74 x 0.74 ................(Assuming that the seating process is independent)

= 0.405

B) Probability that only 1 person in the group of 3 students is left-handed

(L,R,R) or (R,L,R) or (R,R,L)

= Pr( Left handed person seated at the extreme left) + Pr(Left handed person seated in the middle) +Left handed person seated at the extreme right)

= P(L)xP(L')xP(L') + P(L')xP(L)xP(L') + P(L')xP(L')xP(L)

= 0.26 x 0.74 x 0.74 + 0.74 x 0.26 x 0.74 + 0.74 x 0.74 x 0.26

= 0.427

c) There would be no complaints if the left handed person is seated in the far left seat. Also in one table that can seat 3 test takers, if more than one of them is left handed as they would constantly bump elbows with the person next to them causing complaints.

Hence, two cases are feasible here

-  No one in a group of 3 students is left-handed (or)

- One person is left handed and is seated in the extreme left.

Hence, the required probability would be:

0.405 + 0.142 = 0.548

D) Probability of no complaint

(L,R,R) (L,R,R) (R,R,R) or(L,R,R) (R,R,R) (L,R,R) or (R,R,R) (L,R,R) (L,R,R)

= Pr (Both left handed people are seater at the extreme left of first two tables) + Pr (Both left handed people are seater at the extreme left of first and third table) + Pr (Both left handed people are seater at the extreme left of last two tables )

= [(0.26 x 0.742)(0.26 x 0.742)(0.743)] + [(0.26 x 0.742)(0.743)(0.26 x 0.742)] + [(0.743)(0.26 x 0.742)(0.26 x 0.742)]

= 0.0246