MCQ Test: Probability- 2 - Bank Exams MCQ

Total characters = 9, vowels = 2, consonants = 7

Except the end places, vowels can be arranged at 7 places

No of ways = ⁷C₂ – 6 (minus 6 for the chances when both vowels are together) = 15

No of ways of arranging the 7 consonants

Letters are K – 3, A – 2, C – 2, R, and J

Hence, option D is correct.

A team has six girls and six boys. Three students have to be selected for a project.

Calculation:

⇒ Total number of students = 12

⇒ Probability that three students are selected = ¹²C₃ = 12 x 11 x 10/6 = 220

⇒ Probability that two girls and one boy are selected = ⁶C₂ x ⁶C₁ = 6 x 5/2 x 6 = 90

∴ Required probability = 90/220 = 9/22

Total number of ways to select team blue without any restriction = ⁹C₅ Similarly team Red can be selected in ⁹C₅ ways

Total number of ways to select both the teams = ⁹C₅ x ⁹C₅

P (at least one of them plays) = 1 – P (none of them plays)

Total number of ways of selecting team without selecting ankit and vaibhav = ⁸C₅ x ⁸C₅

P (at least one of them plays)

Hence, option B is correct.

Pages = 1 cover page, 12 theory pages, 6 pictures page

Except cover page

Ways of arranging 12 + 6 pages = 18!

Ways of arranging so that the theory pages are in order and drawing pages come together= 13C1 × 6!

(As there are 13 gaps between 12 pages where 6 pages can be kept)

Hence, option C is correct.

GIVEN:

A bag contains coins of one rupee, two rupee & five rupees.

The total money in the bag is Rs. 120.

The total number of one rupee and two rupee coins are 35 and in ratio of coins is 2 : 5.

Formula used:
Probability of an event = Favorable ways/Total ways

CALCULATION:

Number of 1 rupee coins = (2/7 × 35) = 10

Number of 2 rupee coins = (5/7 × 35) = 25

Let the number of 5 rupee coins be x

5x + 1(10) + 2(25) = 120 ⇒ x = 12

Total number of coins = 10 + 25 + 12 = 47

So the bag contain total 47 coins out of which number of five rupee coin is 12

Probability of getting a 5-rupee coin = 12/47

Given:

Probability of red ball = 1/3

Probability of white ball = 1/2

Formula Used:

Probability = (Number of successful outcomes/Total number of outcomes)

P(E) = (nE)/(nS), where nE = Number of events and nS = Number of sample space 

Calculation:

Probability of getting green ball = 1 - (1/3 + 1/2)

⇒ 1 - 5/6 = 1/6

According to the question:

If one unit corresponds to 9 green balls then,

6 unit = 6 × 9 = 54

Total no. Of balls = 54

∴ The total number of balls in the bag is 54.

Since one die can be thrown in six ways to obtain any one of the six numbers marked on its six faces

⇒ Total number of elementary events = 6 x 6 x 6 = 216

Let A be the event of getting a total of at least 6.Then Ā denotes the event of getting a total of less than 6 i.e. 3, 4, 5.

⇒ Ā = { (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3),(1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1) }

So, favorable number of cases = 10

Hence, option (C) is correct.

Given:

Three coins are tossed simultaneously.

Formula:

Probability = Number of favorable outcomes/ Total number of outcomes.

Calculation:

When three coins are tossed then the outcome will be any one of these combinations. (TTT, THT, TTH, THH. HTT, HHT, HTH, HHH).

So, the total number of outcomes is 8.

Now, for exactly two heads, the favorable outcome is (THH, HHT, HTH).

We can say that the total number of favorable outcomes is 3.

Again, from the formula

Probability = Number of favorable outcomes/Total number of outcomes

Probability = 3/8

∴ The probability of getting exactly two heads is 3/8.

Given

Number of black balls = 5

Number of white balls = 6

Formula

Probability = Favorable events/Total possible events

Calculation

Favorable event = ⁶C₂

Total possible events = ¹¹C₂

∴ Probability = ⁶C₂/¹¹C₂ = (6 x 5)/(11 x 10) = 3/11

Correct Option: C
Total number of ways of selecting 3 students from 25 students = ²⁵C₃

Number of ways of selecting 1 girl and 2 boys = selecting 2 boys from 15 boys and 1 girl from 10 girls

⇒ Number of ways in which this can be done = ¹⁵C₂ x ¹⁰C₁

Hence, option C is correct.

If 1 apple is rotten + 2 apples are other

= ⁴C₁ x ⁶C₂ = 60

If 2 apples are rotten + 1 apple is other

= ⁴C₂ x ⁶C₁ = 36

If 3 apples are rotten

= ⁴C₃ = 4

Total outcomes = ¹⁰C₃ = 120

Hence, option B is correct.

Let the names of children be x, y and z. The probabilities of the three children to finish the race are 1/3, 1/5 and 1/4 respectively. It may be noted that one reaching the finishing point is independent of other reaching. If P(x), P(y) and P(z) denotes the probabilities.

The probability of at least one of them reaching the finishing point = 1 – P (none of them finishing the race)

Hence, option B is correct.

Hence, option B is correct.

Hence, option D is correct.

Let the tiffin boxes be T1 and T2. T1 contains x pink and x-4 yellow toffees; T2 contains x-3 pink and x-1 yellow toffees. The possibility is that either of T1 or T2 is selected with a probability of 1/2 in each case. Having selected a tiffin, two different toffees are selected. The probability is 67/132.

Total number of toffees in first tiffin = x + x – 4 = 2x – 4

Now putting the value of total number of toffees i.e.; 2x – 4 in (1)using options

From option (a)

2x – 4 = 8

x = 6

put value of x in (1)

From option (b)

2x – 4 = 12

x = 8

put value of x in (1)

Hence, option B is correct.