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

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 \(\dfrac{^4C_1 \times \ ^2C_1}{^9C_2}\)

Probability of choosing a white ball then a red ball from bag \(A\) is given by \(\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}\)

So, the probability of getting a white ball and then a red ball from bag \(B\) is given by = \(\dfrac{\dfrac{^4C_1 \times \ ^2C_1 }{^9C_2}}{\dfrac{^4C_1 \times ^2C_1}{^9C_2}+\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}}\)

\(\Rightarrow\dfrac{\dfrac{2}{9}}{\dfrac{2}{7}+\dfrac{2}{9}} = \dfrac{2 \times 7}{18+14}=\dfrac{7}{16}\)

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 \(\dfrac{^4C_1 \times \ ^2C_1}{^9C_2}\)

Probability of choosing a white ball then a red ball from bag \(A\) is given by \(\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}\)

So, the probability of getting a white ball and then a red ball from bag \(B\) is given by = \(\dfrac{\dfrac{^4C_1 \times \ ^2C_1 }{^9C_2}}{\dfrac{^4C_1 \times ^2C_1}{^9C_2}+\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}}\)

\(\Rightarrow\dfrac{\dfrac{2}{9}}{\dfrac{2}{7}+\dfrac{2}{9}} = \dfrac{2 \times 7}{18+14}=\dfrac{7}{16}\)

Hence option 1 is correct

Explanation:

Possible outcomes x < y < z

= = (1, 2, 3) (1, 2, 4), (1, 2, 5), (1, 2, 6) (1, 3, 4) (1, 3, 5) (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4),(2, 3, 5) (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6) (3, 5, 6), (4, 5, 6)

Thus, Total possible outcomes = 20

∴ Option (a) is correct.

Explanation:

⇒n(S) = 6 × 5 × 4

⇒ n(E) = n (one of the faces shown is an ace.)

=  3(1 × 5 × 4) = 3 × 5 × 4 

⇒ \(P(E) = \frac{3\times 5\times 4}{6\times 5\times4} = \frac{1}{2}\)

∴ Option (d) is correct

Explanation:

⇒n(S) = 6 × 6 = 36

⇒Sum 9 = (3, 6), (4, 5), (5, 4), (6, 3)

⇒ Sum 10 = (4, 6), (5, 5), (6, 4)

⇒P (sum 9 or 10) = 7/36

Then, P (sum neither 9 nor 10) = \(1- \frac{7}{36} =\frac{29}{36}\)

∴ Option (d) is correct

Concept:

• The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as: \(\rm P(k) =\dfrac{^{p}C_{k}}{^{n}C_{k}}\).
• ​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 = \(\rm \dfrac{^{7}C_{1}}{^{11}C_{1}}=\dfrac{7}{11}\).

Probability of drawing 1 blue ball = \(\rm \dfrac{^{4}C_{1}}{^{11}C_{1}}=\dfrac{4}{11}\).

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

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

Probability of getting the balls of different colors = \(\dfrac{28}{121}\) + \(\dfrac{28}{121}\) = \(\dfrac{56}{121}\)

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 = \(\frac{{{\rm{Number\;of\;ways\;it\;can\;happen}}}}{{{\rm{Total\;number\;of\;outcomes}}}}\)

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 = \(\frac{{{{\rm{\;}}^8}{{\rm{c}}_2}}}{{{{\rm{\;}}^{16}}{{\rm{c}}_4}}}\)

Concept:

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

\(\rm P(E)=\frac{n(E)}{n(S)} \)

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 = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)

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 = \(\rm \text{Total favorable cases}\over\text{Total possible cases}\)

P = \(3\over4\)

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 = \(\rm \dfrac{\text{(Number of ways it can happen)}}{\text{ (Total number of outcomes)}}\)

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) = \(26\over 52\)

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

So the probability of getting either of them (P₂) = \(2\over 52\)

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

P = \(26+2\over 52\)

P = \({28\over 52} = \boldsymbol{7\over 13}\)

Concept:

The probability of the occurrence of an event A, out of total possible outcomes N, is given by P(A) = \(\rm \dfrac{n(A)}{N}\), 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 = \(\rm \dfrac{n(A)}{N}=\dfrac{2}{8}=\dfrac{1}{4}\)

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 = \(\rm \dfrac{n(A)}{N}=\dfrac{2}{8}=\dfrac{1}{4}\)