Test: Probability And Expected Value By Mathematical Expectation- 6 - CA Foundation MCQ

Total events are 6
Possibility of Greater than 2 is 4 i.e. 3, 4, 5, 6
Hence, Probability of the occurrence of a no. greater then 2 will be 4/6

=2/3

• Number of red cards = 5

• Number of red kings = 2

• Probability (king | red) = 2/5

• P(Biology) = 0.25

• P(Mathematics and Biology) = 0.15

• P(Mathematics given Biology) = 0.15 / 0.25 = 3/5

Answer: B: 3/5

• P(Math) = 0.40

• P(Bio and Math) = 0.15

• P(Bio given Math) = 0.15 ÷ 0.40 = 3/8

Answer: C: 3/8

The odd faces on a six-sided die are {1, 3, 5}, three out of six faces.

So the probability is 3/6=1/2

Answer: A: 1/2

For a certain A, P(A) is 1

When no outcomes are favourable to the event, the event is impossible.

 impossible

• P(A and B) = 0.25

• P(B) = 0.35

• P(A given B) = 0.25 ÷ 0.35 = 5/7 

  ≈ 0.7143

in a pack of card there are total  52 Cards

Spade cards  are   13  

Probability that two consecutive cards drawn  from a well shuffled pack of cards are spade = ¹³C₂ / ⁵²C₂  

= 13 * 12 / 52 * 51

= 12 / 4 * 51

= 3/51

= 1/17

other ways

probability of Choosing 1st card as spade = (13/52) = 1/4

Probability of choosing 2nd spade card = 12/51  = 4/17

(1/4)(4/17)  = 1/17

Step 1: Total Possible Outcomes
The total number of outcomes when throwing two dice is:
6 × 6 = 36

Step 2: Favorable Outcomes

• Sum = 7: The combinations for a sum of 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Total = 6 outcomes.
• Sum = 11: The combinations for a sum of 11 are: (5,6), (6,5). Total = 2 outcomes.

The total number of outcomes where the sum is either 7 or 11:
6 + 2 = 8

Step 3: Outcomes Where the Sum is Neither 7 nor 11
The outcomes where the sum is neither 7 nor 11 are:
36 - 8 = 28

Step 4: Probability
The probability is the ratio of favorable outcomes to total outcomes:
P = 28 / 36 = 7 / 9

The probability that the sum is neither 7 nor 11 is 7/9.

When two coins are tossed simultaneously, the possible outcomes are:

• HH (both heads)

• HT (head on the first coin, tail on the second)

• TH (tail on the first coin, head on the second)

• TT (both tails)

Out of these, the outcomes with exactly one head are HT and TH.

There are 4 possible outcomes in total, and 2 of them have exactly one head.

So, the probability of getting exactly one head is: 2/4 = 1/2

Thus, the correct answer is:

d) 1/2.

Check that these probabilities sum to 1:
3/20 + 1/5 + 1/2 + 1/10 + 1/20 = 1

Calculate the expected value E(X)

E(X) = (-20)(3/20) + (-10)(1/5) + 30(1/2) + 75(1/10) + 80(1/20)
= -3 + -2 + 15 + 7.5 + 4
= -5 + 15 + 7.5 + 4
= 21.5

Hence, the expected value of X is 21.5.

Let's denote:

• U₁: the event of selecting urn 1
• U₂: the event of selecting urn 2
• R: the event that a red ball is drawn

The probability of selecting either urn is ½.

For urn 1 (U₁):
It has 2 red and 1 green ball, so P(R | U₁) = 2/3.

For urn 2 (U₂):
It has 2 red and 2 green balls, so P(R | U₂) = 2/4 = 1/2.

Now, using Bayes’ theorem, we find the probability that the red ball came from urn 1:

  P(U₁ | R) = [P(R | U₁) × P(U₁)] / [P(R | U₁) × P(U₁) + P(R | U₂) × P(U₂)]

Substitute the values:

P(U₁ | R) = [(2/3) × (1/2)] / [(2/3) × (1/2) + (1/2) × (1/2)]         
= (1/3) / [(1/3) + (1/4)]         
= (1/3) / [(4/12) + (3/12)]         
= (1/3) / (7/12)         
= (1/3) × (12/7)         
= 12/21         
= 4/7

Thus, the probability that the red ball was drawn from urn one is 4/7.

Correct answer: (C) 4/7

 In a packet of 500 pens, there are 50 defective pens. Therefore, the number of non-defective pens is 500 - 50 = 450. The probability of selecting a non-defective pen is the number of non-defective pens divided by the total number of pens, which is 450 / 500 = 9/10.

To solve the problem of finding the probability that the selected persons are a gentleman and a lady but not a couple, follow these steps:

• There are 8 people in total: 4 gentlemen and 4 ladies.

• The total number of ways to select any 2 people from these 8 is calculated as:

  C(8, 2) = 28 ways.

• We want a gentleman and a lady who are not a couple. First, count all ways to select one gentleman and one lady:

  4 gentlemen x 4 ladies = 16 ways.

• However, 4 of these pairs are couples (each gentleman with his wife), so we must exclude these:

  16 - 4 = 12 pairs that are not couples.

• Thus, the probability is:

  12/28 = 3/7.