MCQ Test: Experimental Probability - 1 - Banking Exams MCQ

We can see that total number of events of being 35-50 years of age and having more than 1 accidents = 145
Total number of events = 1500
Hence, probability of being 18-29 years of age and having more than 1 accidents = 145/1500
= 0.29.

We can see that total number of events of being elder than 35 years of age and having at least 1 accident = 277 + 78 + 118 + 59
= 532
Total number of events = 1500
Hence, probability of being 18-29 years of age and having more than 1 accidents = 532/1500
= 0.354.

We can see that total number of events of being 18-35 years of age and having more than 1 accidents = 90
Total number of events = 1500
Hence, probability of being 18-35 years of age and having more than 1 accidents = 90/1500
= 0.06.

In this case, having at least one boy means one boy or two boys.
Hence, number of families having at least one boy = 415 + 315 = 730
Therefore, the probability that the family has at least one boy  = 0.73

It can be seen that the student have scored 3 out of 5 times more than 80 marks.
Therefore, probability that the student has scored more than 80 = 
= 3/5
= 0.6.

We can see from the table that we get ‘4’ 79 times out of 500 trials.
Therefore, probability of getting ‘4’ as outcome = 
= 79/500
= 0.158.

Number of events of obtaining minimum one head = 55 + 105
= 160
Hence, P (E) = probability of event of obtaining minimum on head

= 160/200
= 0.8.

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) = 26/52

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

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

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

P = 28/52 = 7/13

Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

Calculation:

Two dice thrown,

S = 

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

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

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

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

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

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

n(S) = 36

Now, let A be an event getting sum 7

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

n(A) = 6,

∴Probability of getting sum 7 = n(A) /n(S)

= 6/36

= 1/6

Hence, option (a) is correct.

Concept:

The probability of the occurrence of an event A, out of total possible outcomes N, is given by P(A) = 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 =

Concept:

Probability of an event happening = 

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 (a) is correct.

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: 

• The number of ways for selecting r from a group of n (n > r) = nCr 
• The probability of particular case =

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 = 
P = 3/4

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 
• 

2) Probability of an event happening = 

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 = 

Concept:

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

P(E) = n(E)/n(S)

Calculation:

Total fruits = 3 + 3 = 6

Total possible ways = ⁶C₂ = 15 = n(S)

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

∴ Required probability = 9/15 = 3/5

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:

• The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as: 
• ​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 =

Probability of drawing 1 blue ball = 

Probability of drawing (1 red) AND (1 blue) ball = 

Similarly, Probability of drawing (1 blue) AND (1 red) ball = 

Probability of getting the balls of different colors = 

Key Points

• The probability of rolling a sum of 7 on two fair dice is 5/36
• The probability is calculated by finding the number of favorable outcomes (rolling a sum of 7) divided by the total number of possible outcomes (rolling two dice).
• The total number of possible outcomes when rolling two dice is 6 x 6 = 36
• The favorable outcomes that result in a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1)

Additional Information

• Probability: the measure of the likelihood of an event occurring, expressed as a number between 0 and 1
• Fair dice:
  • a die that is not biased and has an equal chance of landing on any number from 1 to 6
• Mutually exclusive events:
  • events that cannot happen at the same time
• Complementary events:
  • events that are mutually exclusive and add up to 1.
• Independent events:
  • events that the outcome of one does not affect the outcome of the other.
• Mutually exclusive and independent events are different.

Key Points

•  When events A and B are independent, the outcome of one event does not affect the outcome of the other event. 
• Therefore, the probability of A and B happening is the product of the individual probabilities of A and B.
• The probability of A and B happening is P(A) * P(B)  when events A and B are independent
  • the outcome of one event does not affect the outcome of the other event.

Additional Information

To solve this problem, we can use the formula for the probability of independent events:

               P(A and B) = P(A) * P(B)  

To solve this problem, we can use the formula for the probability of independent events: P(A and B) = P(A) * P(B)  

• Step 1:
  • Find the probability of event A happening: P(A) 
• Step 2:
  • Find the probability of event B happening: P(B)  
• Step 3:
  • Multiply the two probabilities together: P(A and B) = P(A) * P(B)  

Therefore, the answer is P(A) * P(B)

Calculation:

• The probability of getting a tail on each toss is equal to 1 minus the probability of getting a head.
• Given that the probability of getting a head is 1/4, the probability of getting a tail on each toss is 1 - 1/4 = 3/4.
• Since each toss is independent, the probability of getting a tail on the first four tosses followed by a head on the fifth toss can be calculated by multiplying the probabilities of each individual toss.

Therefore, the probability of getting a tail in all the first four tosses followed by a head is:

(3/4) × (3/4) × (3/4) × (3/4) × (1/4) = (3/4)⁴× (1/4)

Simplifying the expression:

(3/4)⁴× (1/4) = 81/256 × 1/4 = 81/1024

Therefore, the probability of getting a tail in all the first four tosses followed by a head is 81/1024.

The correct answer is option (b)