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

Solution:

• Probability of tie: Occurs when both A and B get equal heads.
  • (A = 0, B = 0): 1/4 × 1/8 = 1/32
  • (A = 1, B = 1): 1/2 × 3/8 = 3/16
  • (A = 2, B = 2): 1/4 × 3/8 = 3/32
  • Total tie probability T = 1/32 + 3/16 + 3/32 = 5/16
• Let E = expected number of rounds until someone wins
• The recurrence is: E = 1 + T × E
• ⇒ E − (5/16)E = 1
• ⇒ (11/16)E = 1
• ⇒ E = 16/11

∴ Final Answer: 11 E = 16 

Solution:

• A tosses 2 coins ⇒ Heads: 0, 1, 2 with probabilities: 1/4, 1/2, 1/4
• B tosses 3 coins ⇒ Heads: 0, 1, 2, 3 with probabilities: 1/8, 3/8, 3/8, 1/8

Favorable outcomes for A:

• (1, 0): 1/2 × 1/8 = 1/16
• (2, 0): 1/4 × 1/8 = 1/32
• (2, 1): 1/4 × 3/8 = 3/32

Total P(A wins) = 1/16 + 1/32 + 3/32 = 3/16

P(B wins): = 1/2

P(Tie): = 1 − (3/16 + 1/2) = 5/16

Let P = final probability that A wins

⇒ P = 3/16 + 5/16 × P

⇒ P − (5/16)P = 3/16

⇒ (11/16)P = 3/16

⇒ P = 3/11

∴ K = 3

Concept:

• P(A') = 1 - P(A)
• P(A' ∩ B') = P((A ∪ B)') {Probability of not A and not B}
• P(A and not B) = P(A ∩ B') = P(A) - P(A ∩ B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given:  P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.3, 

⇒  P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒  P(A ∪ B) = 0.5 + 0.7 - 0.3 = 0.9__(i)

Now the probability of not A and not B,

P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B) 

⇒ P(A' ∩ B') = 1 - 0.9 { From (i)}

⇒ P(A' ∩ B') = 0.1 __(ii)

The probability of B and not A,

P(A' ∩ B) = P(B) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.7 - 0.3 = 0.4 __(iii)

The probability of A and not B,

P(A ∩ B') = P(A) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.5 - 0.3 = 0.2 __(iv)

From (ii), (iii) and (iv),

P(A' ∩ B') + P(A' ∩ B) + P(A ∩ B') = 0.1 + 0.4 + 0.2 = 0.7

∴ The correct option is (2).

Explanation -

x + y = 24, x, y ∈ N 

AM > GM

⇒ xy ≤ 144

xy ≤ 108 

Favorable pairs of (x, y) are

(13, 11), (12, 12), (14, 10), (15, 9), (16, 8),

(17, 7), (18, 6), (6, 18), (7, 17), (8, 16), (9, 15),

(10, 14), (11, 13)

i.e. 13 cases 

Total choices for x + y = 24 is 23

Probability = \(\frac{13}{23}\) = \(\frac{\mathrm{m}}{\mathrm{n}}\)

n – m = 10  

Hence Option (4) is correct.

Given:

P(A) = (3)/(8)

P(B) = (1)/(2)

P(A ∩ B) = (1)/(4)

Formula used:

P(A̅ ∩ B̅) = 1 - P(A ∪ B)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

P(A ∪ B) = (3)/(8) + (1)/(2) - (1)/(4)

⇒ P(A ∪ B) = (3)/(8) + (4)/(8) - (2)/(8)

⇒ P(A ∪ B) = (5)/(8)

⇒ P(A̅ ∩ B̅) = 1 - (5)/(8)

⇒ P(A̅ ∩ B̅) = (3)/(8)

∴ 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: \(\rm P(k) =\dfrac{^{p}C_{k}}{^{n}C_{k}}\).
• ​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:

• For two events A and B, we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).

Calculation:

Using the concept above, because A and B are independent events, we can write:

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.7 = 0.4 + P - 0.4 × P

⇒ 0.6P =0.3

⇒ P = 0.5.

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:

• \(\rm P(A|B) = \frac {P(A \;∩ \; B)}{P(B)}\)
• \(\rm P(B|A) = \frac {P(A \;∩ \; B)}{P(A)}\)
• A ⊂ B = Proper Subset: every element of A is in B, but B has more elements.
• ϕ = Empty set = {}

Calculation:

Given: P(B/A) = 1

⇒ \(\rm P(B|A) = \frac {P(A \;∩ \; B)}{P(A)} = 1\)

⇒ P(A ∩ B) = P(A)

⇒ (A ∩ B) = A

So, every element of A is in B, but B has more elements.

∴ A ⊂ B

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:

Permutations with Repetition = nr

Where n is the number of things to choose from r different things when repetition is allowed, and order matters.

Favorable cases = Total cases - Unfavorable cases

Calculation:

According to the question

Four dies are rolled 

So, Total Possible number of outcomes = 64

Now, Total outcomes when no 2 appears = 54

Now, From the concept used

Favorable cases = 64 - 54

⇒ 1296 - 625

⇒  671

∴ The number of possible outcomes in which at least one die shows 2 is 671.

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