Test: Conditional Probability - Electronics and Communication Engineering (ECE) MCQ

We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
Which is equivalent to (3/13) / (7/13), hence the value of P(A|B) = 3/7.

We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
Also P(A∪B) = P(A)+P(B) – P(A∩B). (By formula of probability)
⇒ 8/11 = 7/11 + 6/11 – P(A∩B)
⇒ P(A∩B) = 13/11 – 8/11
⇒ P(A∩B) = 5/11
P(A|B) = (5/11) / (6/11).
P(A|B) = 5/6

We know that P(F|E) = P(E∩F) / P(E). (By formula for conditional probability)
Value of P(E∩F) is given to be 0.3 and value of P(E) is given to be 0.5.
P(F|E) = (0.3) / (0.5).
P(F|E) = 3 / 5.

We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
The value of P(B) = 0 in the given question. As the value of denominator becomes 0, the value of P(A|B) becomes un-defined.

We know that P(S|F) = P(S∩F) / P(F). (By formula for conditional probability)
Which is equivalent to P(F|F) = P(F) / P(F) = 1, hence the value of P(F|F) = 1.

We know that P(E|F) = P(E∩F) / P(F). (By formula for conditional probability)
Value of P(E∩F) is given to be 0.2 and value of P(F) is given to be 0.3.
P(E|F) = (0.2) / (0.3).
P(E|F) = 2 / 3.

E and F are two events associated with the same sample space of a random experiment.
The value of P (E|F) = (E∩F) / P(F), provided P(F) ≠ 0. We know that if P(F) = 0, then the value of P(E|F) will reach a value which is not defined hence it is wrong option. Also, P(E∩F) / P(F) and P(E∩F) / P(E) are wrong and do not equate to P(E|F).

Let's use conditional probability here. We want to find the probability of selecting Bag A given that a red marble was chosen.

P(A) is the probability of choosing Bag A, and P(R) is the probability of choosing a red marble.

We know that P(A) is 1/2 because we choose a bag at random, and there are two bags.

P(R|A) is the probability of choosing a red marble from Bag A, which is 3/5.

P(R|B) is the probability of choosing a red marble from Bag B, which is 4/5.

Now, we can use Bayes' theorem to find P(A|R), the probability of choosing Bag A given a red marble was picked:

P(A|R) = [P(R|A) * P(A)] / [P(R|A) * P(A) + P(R|B) * P(B)]

P(A|R) = (3/5 * 1/2) / [(3/5 * 1/2) + (4/5 * 1/2)]

P(A|R) = (3/10) / [(3/10) + (4/10)]

P(A|R) = (3/10) / (7/10)

P(A|R) = 3/7

So, the probability of selecting Bag A, given that a red marble was chosen, is 3/7, which corresponds to option D.

To find the probability that the first card is red and the second card is also red, we can use conditional probability.

First, the probability that the first card is red is 26/52 because there are 26 red cards out of 52 in the deck.

Now, after drawing a red card, there are 25 red cards left in a deck of 51 cards (one card has already been drawn).

So, the probability that the second card is also red, given that the first card was red, is 25/51.

Therefore, the probability of both events happening is the product of these two probabilities:

P(First card is red and Second card is red) = (26/52) * (25/51)

Simplifying this:

P(First card is red and Second card is red) = (1/2) * (25/51)

P(First card is red and Second card is red) = 25/102

So, the correct answer is option C, which is 25/102.

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