Question

12. Assume that we randomly choose from the letters {A, B, C, D, E, F, G, H, I, J, K, L} (without replacing the letters), until they have all been taken.

(a) Find the probability that the letters A and K are chosen successively in the given order.

(b) Find the probability that the letters G, H, I, are chosen successively in the given order.

c) Find the probability that the string "LAI" appears somewhere in the sequence of letters.

13. Suppose that a minibus with capacity for 9 passengers departs from a commuter station. Observation has shown that the bus never departs empty. Let X denote the number of passengers that are female and Y the number of passengers that are male. Assume that all possible (x, y) pairs are equally likely. Find the probability that the number of female passengers is less than 5.

Answer 1

12.

total no. of ways to pick = total no. of arrangements of these 12 letters

{total no. of ways to arrangements of n things = n! }

total no. of ways to pick = 12! (! means factorial)

total no. of ways to pick = 479001600

(a) for A and K to be Successive let us Take AK as one object so now we have AK and remaining other 10 letters

so we have to arrange 11 objects

no. of arrangements (AK as one object) = 11!

P(A, K succesive) = no. of arrangements (AK as one object) / total no. of ways to pick

P(A, K succesive) = 11! / 12! = 1/12

P(A, K succesive) = 1/12

(b)for G, H and I to be Successive let us Take GHI as one object so now we have GHI and remaining other 9 letters

so we have to arrange 10 objects

no. of arrangements (GHI as one object) = 10!

P(g, h, i succesive) = no. of arrangements (GHI as one object) / total no. of ways to pick

P(G,H,I succesive) = 10! / 12! = 1/(11*12)

P(G,H,I succesive) = 1/132

(c) string "LAI" appears somewhere in the sequence of letters is same as L,A,I choosen succesively in same sequence

so probability will be same as in (b)

P(LAI appears in the sequence) = 1/132

13.

(x,y)

given that x+y <= 9(maximum capacity) and x+y>0 (bus never departs empty)

no of possible arangements (x,y)

= no of possible arangements (0,y) + no of possible arangements(1,y) + no of possible arangements(2,y) + no of possible arangements(3,y) + no of possible arangements(4,y) + no of possible arangements(5,y) + no of possible arangements(6,y) + no of possible arangements(7,y) + no of possible arangements(8,y) + no of possible arangements(9,y)

no of possible arangements (x,y)

= 9 (if x = 0 y can be from 1 to 9) + 9 (if x = 1 y can be from 0 to 8) + 8 (if x = 2 y can be from 0 to 7) + 7 (if x = 3 y can be from 0 to 6) + 6 (if x = 4 y can be from 0 to 5) + 5 (if x = 5 y can be from 0 to 4) + 4 (if x = 6 y can be from 0 to 3) + 3 (if x = 7 y can be from 0 to 2) + 2(if x = 8 y can be from 0 to 1) + 1 (if x = 9 y can be 0)

= 54

no of arrangements (x<5,y) = 9 (if x = 0 y can be from 1 to 9) + 9 (if x = 1 y can be from 0 to 8) + 8 (if x = 2 y can be from 0 to 7) + 7 (if x = 3 y can be from 0 to 6) + 6 (if x = 4 y can be from 0 to 5)

no of arrangements (x<5,y) = 9+9+8+7+6 = 39

probability that the number of female passengers is less than 5 = P(x<5)

P(x<5) = no of arrangements (x<5,y) / no of possible arangements (x,y)

=39/54

P.S. (please upvote if you find the answer satisfactory)