Question

4. Provide a valid conclusion to the following "story": If it were Brillig then the Shithy-toves did imble in the wabe Either the slithy-toves did not gimble in the wabe, or the burrow-groves were Mims If the roses were Painted red, then the burrow-groves were not mimsy It was brillig 5 5 Russian, 4 English and 6 Chinese books are to be placed on a straight shelf. In how many ways can this be done if a) ) No restrictions are placed on the books? ISI ii) If the Russian books must remain together? If the Russian books are placed first? i) 1,207,174 34 21 5 1D iii) S 10 b) What is the probability that the books will be placed in theb with the books of the same language being placed together?3! S4 6. There are three people: A, B, and C. On a certain issue, p( A is correct)-0.3, P(B is corroct) 0.2 and p(C is correct)-0.4 A question about this issue was asked to the three: What is the probability that: a) all three were correct on the issue? b) exactly one person was correct on the issue? e) If exactly one person was correct, that it was B? a) b) c) A machine makes widgets. 10% of all widgets produced by this machine are defective. 8 widgets are selected at random from the production linc. What is the probability that 3 of them are defective? 8. a) A coin is weighted so that p(Hn)-4/7 The coin is tossod a) 10 times. What is the probability of getting 4 heads? b) In the general population, 20% of the pople have trait A, b) 30% have trait B and the rest have trait C, 20 people are chosen at random. What is the probability that 8 have A, 7 have B and the rest have C

Answer 1

5.

a)

i) Total books =5+4+6= 15. Number of ways of arranging these 15 books =15! =1,307,674,368,000

ii) Russian boos must remain together. Treating Russian books as one unit, we have 1 unit of Russian books+4+6 =11 elements which can be arranged in 11! ways. Again 5 Russian books can be arranged among themselves in 5! ways. Therefore, total number of ways of arranging books so that Russian books must remain together = 11!(5!) =4,790,016,000

iii) If Russian books are placed first, then they can be arranged in 5! ways. Remaining 10 books can be arranged in 10! ways. Therefore, total number of ways of arranging books so that Russian books are placed first = 5!(10!)= 435,456,000

b)

Number of ways of arranging books so that the books of same language are placed together:

Treating 3 sets of books as 3 units, number of ways of arranging these 3 units = 3!

Now 5 Russian books can be arranged among themselves in 5! ways.4 English books can be arranged in 4! ways. 6 Chinese books can be arranged in 6! ways.

Therefore, total number of ways of arranging books so that the books of same language are placed together =3!*5!*4!*6! =6*120*24*720 =12,441,600

Hence, the probability of placing the books with the same language books remaining together =number of ways of placing books with same language books staying together/total number of ways of placing all books =12,441,600/1,307,674,368,000 = 0.0000095142 which is almost 0.

6.

A, B and C being correct on the issue were assumed to be independent events.

a)

The probability of all 3 were correct =0.3*0.2*0.4 =0.024

b)

The probability of exactly one person was correct =0.3(0.8)(0.6)+0.2(0.7)(0.6)+0.4(0.7)(0.8) =0.144+0.084+0.224 =0.452

c)

P(B/exactly one person was correct)=P(exactly B was correct)=P(only B was correct)=0.2(0.7)(0.6) = 0.084

7.

Binomial distribution:

Given: sample size, n =8; r =3; probability of success on single trial, p =10% =0.10

Formula: P(X=r) =nC_r p^r (1-p)^n-r

P(X=3) =8C₃ (0.1)³ (0.9)⁵ =56(0.001)(0.59049) =0.0331