C) students will speak in the following D) A or B wll speak first order: DECABF

Question

C) students will speak in the following D) A or B wll speak first order: DECABF lo play the lottery in a certain state, a person has to correctly select 5 out of 45 numbers, paying $1 r each five-number selection. If the five numbers picked are the same as the ones drawn by the ttery, an enormous sum of money is bestowed. A) a person with one ticket will win ? B) a person with 100 different tickers will win? committee consisting of 6 people is to be selected from eight parents and four teachers. Find the 12c0-924 bability of selecting B) no teachers A) three parents and three teachers &a; c3. 4c3 224 224/a2433 -ab D) not the same number of teachers and parents

Explanation / Answer

Solution

Back-up Theory

Number of ways of arranging n things among themselves = n!

= n(n - 1)(n - 2) …… 3.2.1………………………………………………………..….(1)

Number of ways of selecting r things out of n things is given by nCr

= (n!)/{(r!)(n - r)!}………………………………………………………………….…(2)

Probability of an event E, denoted by P(E) = n/N ………………………………..…(3)

where n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and N = n(S) = Total number all possible outcomes/cases/possibilities.

Now, to work out the answer,

Part (a) Speaking order of 6 students

Without any restrictions, order of speaking for 6 students can be done in 6! = 720 ways.

C) DECABF is just one of these 720 orders. Hence, P(DECABF) = 1/720 ANSWER 1

D) If first speaker has to be A or B, this can be done in 2 ways. Then, the remaining 5 can be in any order in 5! = 120 ways.

Thus, P(A or B speaks first) = (2 x 120)/720 = 1/3 ANSWER 2

Part (b) Lottery Problem

5 out 45 tickets can be selected in 45C5 = 1221759.

A) P(1 ticket winning) = 1/1221759 ANSWER 3

B) P(100 tickets winning) = 100/1221759 ANSWER 4

Part (c) Committee Formation

A committee of 6 out of 12 (8 parents and 4 teachers) can be formed in 12C6 = 924 ways

A) 3 out of 8 parents can be chosen in 8C3 = 56 ways and

    3 out of 4 teachers can be chosen in 4C3 = 4 ways.

So, P(committee has 3 parents and 3 teachers) = (56 x 4)/924 = 0.2424 ANSWER 5

B) No teacher => all 6 are parents. 6 out of 8 parents can be chosen in 8C6 = 28 ways So, P(committee has no teachers) = 28/924 = 0.0303 ANSWER 6

DONE

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