Maths Probability Test 18: Mathematics of Chance

Explanation:

Let E1 be the event of drawing a red card.

Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13 = 1/26

Explanation:

P(at least one graduate) = 1 – P(no graduates) =

1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6

Explanation:

If both agree stating the same fact, either both of them speak truth of both speak false.

Probability = 3/5 * 4/7 + 2/5 * 3/7
= 12/35 + 6/35 = 18/35

Explanation:

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)

As remaining person can be any one among three couples left.
Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅
= (10 * 6)/252 = 5/21

Explanation:

P(winning prize atleast on one ticket)

= 1 – P(“Losing on all tickets”)
= 1 – (0.8)4 = (1 + (0.8)2)(1 – (0.8)2)
= (1.64)(0.36) = 0.5904

Explanation:

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5

Explanation:

A leap year has 52 weeks and two days

Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = 1/7

Explanation:

There is no definite formula for finding prime numbers among 15 consecutive numbers. Hence the probability cannot be determined.

Explanation:

Two balls can be picked from nine balls in ⁹C₂ ways.

We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105
Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35
Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 * 35) = 2/21

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is = ¹²C₄/¹⁵C₄ = (12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) = 33/91

Explanation:

Required probability = 1 – 1/126 = 125/126

Explanation:

Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126

Explanation:

Using question number 11 and 12, we get the probability as

1 – (1/36 + 5/9) = 5/12

Explanation:

No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.

Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.