Probability of Random Experiments MCQ Quiz - Objective Question with Answer for Probability of Random Experiments - Download Free PDF

Given, P(A) = 2P(B) = 3P(C)

⇒ P(C) = 2/3 P(B)

Since A, B, C are three mutually exclusive and exhaustive events

∴ P(A) + P(B) + P(C) = 1

⇒ P(B) = 3/11

From the given relation P(A) = 2 P(B) = 6/11

Calculation:

Given,

Two fair six-sided dice are rolled.

Each die has faces numbered 1 through 6.

Total number of outcomes:

Since each die has 6 faces, the sample space size is

$N=6\times 6=36$.

Favourable outcomes (sum > 7):

We list all ordered pairs ((i, j)) with (i + j > 7):

Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes

Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Sum = 11: (5,6), (6,5) → 2 outcomes

Sum = 12: (6,6) → 1 outcome

Total favourable outcomes = (5 + 4 + 3 + 2 + 1 = 15).

Probability calculation:

The probability that the sum is strictly greater than 7 is

$P(\text{sum}>7)=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{15}{36}=\frac{5}{12}$.

Hence, the correct answer is Option 2.

Calculation

For the bag $A$ we can see that there are $2$ white, $3$ red and $2$ black balls. Similarly, from bag $B$ we have $4$ white, $2$ red and $3$ black balls.

Probability of choosing a white and then a red ball from bag $B$ is given by ${\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}$

Probability of choosing a white ball then a red ball from bag $A$ is given by ${\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}$

So, the probability of getting a white ball and then a red ball from bag $B$ is given by = ${\displaystyle \frac{{\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}}{{\displaystyle \frac{{}^4{C}_1{\times }^2{C}_1}{{}^9{C}_2}}+{\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}}}$

$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{2}{9}}}{{\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{9}}}}={\displaystyle \frac{2\times 7}{18+14}}={\displaystyle \frac{7}{16}}$

Hence option 1 is correct

Calculation

For the bag $A$ we can see that there are $2$ white, $3$ red and $2$ black balls. Similarly, from bag $B$ we have $4$ white, $2$ red and $3$ black balls.

Probability of choosing a white and then a red ball from bag $B$ is given by ${\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}$

Probability of choosing a white ball then a red ball from bag $A$ is given by ${\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}$

So, the probability of getting a white ball and then a red ball from bag $B$ is given by = ${\displaystyle \frac{{\displaystyle \frac{{}^4{C}_1\times {}^2{C}_1}{{}^9{C}_2}}}{{\displaystyle \frac{{}^4{C}_1{\times }^2{C}_1}{{}^9{C}_2}}+{\displaystyle \frac{{}^2{C}_1\times {}^3{C}_1}{{}^7{C}_2}}}}$

$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{2}{9}}}{{\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{9}}}}={\displaystyle \frac{2\times 7}{18+14}}={\displaystyle \frac{7}{16}}$

Hence option 1 is correct

Given:

• Total red balls = 8

• Total white balls = 5

• Total balls = 8 + 5 = 13

• Number of balls drawn = 3

• Condition: 1 red ball and 2 white balls

Concept Used:

To calculate the probability of an event, we use the formula:

Probability = (Number of favorable outcomes) / (Total possible outcomes)

The total number of ways to choose 3 balls from 13 is given by:

Total combinations: C(13, 3) = 13! / (3! × (13 - 3)!)

For favorable outcomes:

Choose 1 red ball out of 8: C(8, 1) = 8

Choose 2 white balls out of 5: C(5, 2) = 5! / (2! × (5 - 2)!)

Calculation:

Total possible outcomes:

C(13, 3) = 13! / (3! × 10!) = (13 × 12 × 11) / (3 × 2 × 1)

⇒ C(13, 3) = 286

Favorable outcomes:

C(8, 1) = 8

C(5, 2) = 5! / (2! × 3!) = (5 × 4) / (2 × 1)

⇒ C(5, 2) = 10

Total favorable outcomes = C(8, 1) × C(5, 2) = 8 × 10 = 80

Probability:

P = Favorable outcomes / Total outcomes

⇒ P = 80 / 286

Simplify:

⇒ P = 40 / 143

Conclusion:

The probability of drawing 1 red ball and 2 white balls is: 40 / 143

∴ The correct answer is Option 2.

Concept:

• The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as: $\mathrm{P}(\mathrm{k})={\displaystyle \frac{{}^{\mathrm{p}}{\mathrm{C}}_{\mathrm{k}}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{k}}}}$.
• ​Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

  P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

  ('and' means '×' and 'or' means '+')

Calculation:

There are a total of 7 red + 4 blue = 11 balls.

Probability of drawing 1 red ball = ${\displaystyle \frac{{}^7{\mathrm{C}}_1}{{}^{11}{\mathrm{C}}_1}}={\displaystyle \frac{7}{11}}$.

Probability of drawing 1 blue ball = ${\displaystyle \frac{{}^4{\mathrm{C}}_1}{{}^{11}{\mathrm{C}}_1}}={\displaystyle \frac{4}{11}}$.

Probability of drawing (1 red) AND (1 blue) ball = ${\displaystyle \frac{7}{11}}\times {\displaystyle \frac{4}{11}}={\displaystyle \frac{28}{121}}$.

Similarly, Probability of drawing (1 blue) AND (1 red) ball = ${\displaystyle \frac{4}{11}}\times {\displaystyle \frac{7}{11}}={\displaystyle \frac{28}{121}}$.

Probability of getting the balls of different colors = ${\displaystyle \frac{28}{121}}$ + ${\displaystyle \frac{28}{121}}$ = ${\displaystyle \frac{56}{121}}$

Concept:

Independent events:

Two events are independent if the incidence of one event does not affect the probability of the other event.

If A and B are two independent events, then P(A ∩ B) = P(A) × P(B)

Calculation:

Given: P(B) = 0.4 and P(A ∪ B) = 0.6

P(A ∪ B) = 0.6

⇒ P(A) + P(B) - P(A ∩ B) = 0.6

⇒ P(A) + P(B) - P(A) × P(B) = 0.6               (∵ A and B are independent events.)

⇒ P(B) + P(A) [1 - P(B)] = 0.6

⇒ 0.4 + P(A) [1 - 0.4] = 0.6

⇒ P(A) × 0.6 = 0.2 

Concept:

1) Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects is denoted by ${}^n{C}_r$
• ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

2) Probability of an event happening = $\frac{\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{c}\mathrm{a}\mathrm{n}\mathrm{h}\mathrm{a}\mathrm{p}\mathrm{p}\mathrm{e}\mathrm{n}}{\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{s}}$

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

Given:

In a room, there are eight couples.

⇒ Eight couples = 16 peoples

We have to select four peoples out of 16 peoples.

⇒ Total possible cases = ¹⁶C₄

Now, we have to select four people- they may be couples

So, we have to select two couples from eight couples.

⇒ Favourable cases = ⁸C₂

Hence Required Probability = $\frac{{}^8{\mathrm{c}}_2}{{}^{16}{\mathrm{c}}_4}$

Concept:

If S is a sample space and E is a favourable event then the probability of E is given by:

$\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}$

Calculation:

Total fruits = 3 + 3 = 6

Total possible ways = ⁶C₂ = 15 = n(S)

Favourable ways = ³C₁ × ³C₁ = 9 = n(E)

Concept: 

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case = $\frac{\text{Number of ways for the case can be executed}}{\text{Total number of ways for selection}}$

Calculation

If it is known that third toss gets head, the possible cases:

(H, H, H), (H, T, H), (T, H, H), (T, T, H)

∴ Total cases possible = 4

Total favourable cases = 3 [(H, H, H), (H, T, H), (T, H, H)]

So, required probability P = $\frac{\text{Total favorable cases}}{\text{Total possible cases}}$

P = $\frac{3}{4}$

Concept:

P(A) = $\frac{n(A)}{n(S)}$

Where n(A) = No. of favourable cases for event A and n(S) = cardinality of sample space.

Solution:

If a coin is tossed thrice, possible outcomes are:

S = {HHH, HHT, HTH, THH, THT, TTH, HTT, TTT}

Probability of getting one or two heads:

A = {HHT, HTH, THH, THT, TTH, HTT}

$P(A)=\frac{6}{8}$

= $\frac{3}{4}$

Concept: 

Sample space is nothing but a set of all possible outcomes of the experiment.

If we toss a coin n times then possible outcomes or number of elements in sample space = 2ⁿ elements

Calculation:

Number of outcomes when a coin is tossed = 2 (Head or Tail)

∴Total possible outcomes when a coin is tossed 6 times = 2 ×  2 × 2 × 2 × 2 × 2 = 64

Concept:

Probability of an event happening = ${\displaystyle \frac{\text{(Number of ways it can happen)}}{\text{ (Total number of outcomes)}}}$

If a die thrown, Number of sample space = 6, If two dice are thrown n(S) = 6² = 36

Calculation:

Here, four dice are thrown, 

n(S) = 6⁴

Now, sum of the numbers appearing on them 25 = { }       

⇒ n = 0               

(∵maximum sum = 6 + 6 + 6 + 6 = 24)

∴ Probability = 0/(6⁴) = 0

Hence, option (1) is correct.

Concept:

• Either event A alone OR event B alone: m + n.
• Both event A AND event B together: m × n.

Calculation:

There are 26 red cards out of total of 52 cards which also include 2 kings 

So the probability of getting a red card (P₁) = $\frac{26}{52}$

Now from 4 kings as 2 kings are already counted there 2 kings are left

So the probability of getting either of them (P₂) = $\frac{2}{52}$

∴ The probability that the card is red colour or king (P) = P₁ + P₂

P = $\frac{26+2}{52}$

P = $\frac{28}{52}=\text{boldsymbol}\frac{7}{13}$

Concept:

The probability of the occurrence of an event A, out of total possible outcomes N, is given by P(A) = ${\displaystyle \frac{\mathrm{n}(\mathrm{A})}{\mathrm{N}}}$, where n(A) is the number of ways in which the event A can occur.

Calculation:

The total number of different possible outcomes (N) in tossing a coin 3 times is 2³ = 8.

For getting a head and a tail alternately, the possibilities are HTH, THT → 2 possibilities n(A).

∴ Required probability = ${\displaystyle \frac{\mathrm{n}(\mathrm{A})}{\mathrm{N}}}={\displaystyle \frac{2}{8}}={\displaystyle \frac{1}{4}}$

Alternate Method

The possible set of A coin is tossed 3 times is {HHH}{HHT}{HTH}{HTT}{TTT}{TTH}{THT}{THH} = 8

The probability of getting a head and a tail alternately is {HTH}{THT} = 2

So, required probability = ${\displaystyle \frac{\mathrm{n}(\mathrm{A})}{\mathrm{N}}}={\displaystyle \frac{2}{8}}={\displaystyle \frac{1}{4}}$