IC28 Mock Test Sample 20
This section explains probability, counting principles, permutations, and combinations. It covers survival probabilities, independent events, and calculating chances of different outcomes. The multiplication principle states that if one event can happen in several ways and another independently, total ways are obtained by multiplication. Permutations arrange objects in order, while combinations select objects without considering order. Factorials are used in permutation and combination formulas. Circular arrangements and arrangements with repeated objects are also discussed. Examples include arrangements of words, seating guests, travel choices, and survival probabilities. These concepts form the basis for actuarial calculations, statistics, and advanced probability applications.
Q1. In Example 10, probability A and B (both aged 65, P(survive)=.8) and C, D (both aged 70, P(survive)=.7) all alive in 5 years:
a. 0.3136
b. 0.16
c. 0.32
d. 0.42
Q2. In Example 10(b), probability of exactly one death in group I (A and B both 65):
a. 0.16
b. 0.32
c. 0.64
d. 0.04
Q3. In Example 10(c), probability that exactly one person is alive from each group:
a. 0.32
b. 0.42
c. 0.1344
d. 0.3136
Q4. Per the basic principle of counting (5.1 Theorem), if E1 happens in n1 ways and E2 in n2 ways independently:
a. n₁ + n₂
b. n₁ × n₂
c. n₁ / n₂
d. n₁ - n₂
Q5. Per (6.8), the number of permutations of n unlike objects taken r at a time is:
a. n(n−1)(n−2)...(n−r+1)
b. nʳ
c. n!/r!
d. r!
Q6. Per (6.9), n! equals:
a. n(n−1)(n−2)...1
b. nⁿ
c. 1
d. n+1
Q7. Per (6.10), ⁿPₙ equals:
a. n!
b. 1
c. nⁿ
d. 0
Q8. Per (6.11), ⁿPᵣ in factorial notation is:
a. n!/((n−r)!)
b. n!/r!
c. n! × r!
d. r!/n!
Q9. Per (6.12), the value of 0! is:
a. 0
b. 1
c. Undefined
d. Infinity
Q10. Per (6.13), number of circular arrangements of n unlike objects (clockwise/anticlockwise considered different):
a. (n−1)!
b. n!
c. (n+1)!
d. n!/2
Q11. In Example 11 (Andheri to Borivli by bus/rickshaw/train, return likewise), number of ways for to-and-fro journey:
a. 6
b. 9
c. 12
d. 3
Q12. In Example 12, arrangements of letters of “CHILD” beginning with C and ending with L:
a. 120
b. 6
c. 24
d. 12
Q13. In Example 13, 5 guests at a dining table with 2 specific guests together:
a. 120
b. 48
c. 12
d. 24
Q14. Per (6.14), permutations of n objects with p, q, r... alike of each kind:
a. n!/(p! q! r! …)
b. n!
c. n!/(p+q+r)
d. p! q! r! …
Q15. In Example 14, distinct permutations of letters of “COMMITTEE”:
a. 362880
b. 45360
c. 9!/2
d. 181440
Q16. In Example 14(i), arrangements of COMMITTEE with two M's together:
a. 10080
b. 45360
c. 2520
d. 30240
Q17. In Example 14(ii), arrangements with two M's not together:
a. 35280
b. 10080
c. 45360
d. 20160
Q18. In Example 14(iii), arrangements with two M's and two T's together:
a. 10080
b. 2520
c. 5040
d. 2160
Q19. Per (6.15), the number of combinations ⁿCᵣ is:
a. n!/(r!(n−r)!)
b. n!/((n−r)!)
c. n!/r!
d. nʳ
Q20. Per (6.16), ⁿCₙ equals:
a. 1
b. 0
c. n!
d. n
