Probability Test - 2

Total number of cases = 20
Number divisible by 3 and 5 both = 15 (one only)
Therefore, Probability of card with a number divisible by 3 and 5 = 1/20

When a player rolls two dice, there are 6 possibilities for the outcome on each dice.
So, there are 36 possibilities for the outcomes when two dice are rolled.
The possibilities that give a score of 3 or less are:
1 and 1, 1 and 2, 1 and 3, 2 and 1, 2 and 2, 2 and 3, 3 and 1, 3 and 2, 3 and 3.
So, 9 of the 36 possibilities give a score of 3 or less and the probability of getting 3 or less = 9/36 = 1/4
Thus, probability of getting 4 or more = 1 - 1/4 = 3/4

Natural numbers which are divisible by 3 = Integer part of 500/3 = 166
Natural numbers which are divisible by 5 = Integer part of 500/5 = 100
Natural number which are divisible by 15 = Integer part of 500/15 = 33
Natural numbers which are divisible by either 3 or 5 = P(3) + P(5) - P(3 and 5)
= 166 + 100 - 33 = 233
Required probability = 233/500

Probability of solving the problem by Steve P(S) = 1/7
Probability of solving the problem by Jack P(J) = 1/8
Probability that Steve solves the problem or Jack solves the problem = P(S) + P(J) - P(S) . P(J)
= (1/7) + (1/8) - {(1/7) . (1/8)}
= (1/7) + (1/8) - (1/56)
= 14/56 = 1/4

PROBABILITY - Total 11 characters - 4 vowels, 7 consonants
Hence, total number of cases = 11
Favourable cases = 4
Required probability = 4/11

Probability that he succeeds in both the tests = 1 - Probability (does not succeed in any one of the two) = 1 - (0.3 + 0.5 - 0.2) = 0.4

Let E₁, E₂ and E₃ be the events that the problem is solved by A, B and C, respectively. Then,
P(E₁) = 1/3, P(E₂) = 2/7 and P(E₃) = 3/8

Exactly one of A, B and C can solve the problem in the following mutually exclusive ways.
(i) A solves and B and C do not solve.
(ii) B solves and A and C do not solve.
(iii) C solves and A and B do not solve.

Probability that only one will solve the problem
= P(A) and P(not B) and P(not C) + P(not A) and P(B) and P(not C) + P(not A) and P(not B) and P(C)
= (1/3)(5/7)(5/8) + (2/3)(2/7)(5/8) + (2/3)(5/7)(3/8) = 25/56

Total number of possibilities = 6 × 6 × 6 = 216

(i) 6 appearing on all the dice:
There is only one possibility for dice bearing all sixes (6, 6, 6) out of 216 possible outcomes.
Required probability = 1/216

(ii) 4 appearing on at least one die:
This event comprises all the events when 4 appeared on the die, whether once, or twice, or thrice.
Number of outcomes when no 4 appeared = 5 × 5 × 5 = 125
Number of outcomes when 4 appeared on the die at least once = Total number of possibilities - Number of outcomes when no 4 appeared = 216 - 125 = 91
Required probability = 91/216

(iii) Sum of the numbers appearing on the dice to be greater than 3:
Whatever be the possibilities, the least numbers appearing on the dice are (1, 1, 1) and sum = 1 + 1 + 1 = 3; that is, there is only one possibility in which sum is 3.
So, number of outcomes when the sum is greater than 3 = 216 - 1 = 215
Required probability = 215/216

(iv) Sum of the numbers appearing on the dice to be 5:
The possible outcomes of the event are: (1, 1, 3); (1, 2, 2); (1, 3, 1); (2, 1, 2); (2, 2, 1) and (3, 1, 1).
Required probability = 6/216 = 1/36

(i) The probability of an event that is certain to happen is 1. Such an event is called a certain event.
(ii) The sum of the probabilities of all the elementary events of an experiment is 1.
(iii) The event which gives the probability of non occurrence of that event is called its complementary event.