JEE Main 2023 - Probability in One shot | Last 10 Year PYQs | Maths Rapid Revision | Bhoomika Ma'am

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Hi JEE 2023 Aspirants,
PYQs are the best revision quick revisions.
So, in this session, Bhoomika Ma'am will quickly walk you through the Probability questions for JEE Main 2023. Stay Tuned

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[Aakash_JEE]

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[charvi]

Homework:
Q.1. Let E and F be two independent events. The probability that exactly one of them occurs is 11/25 and the probability of none of them occurring is 2/25. If P(T) denotes the probability of occurrence of the event T, then: [IIT JEE 2011]
(A) P(E) = 4/5, P(F) = 3/5
(B) P(E) = 1/5, P(F) = 2/5
(C) P(E) = 2/5, P(F) = 1/5
(D) P(E) = 3/5, P(F) = 4/5
Ans.1. Sorry ma'am, I couldn't solve this question. I tried to write different equations but couldn't reach at the answer.

Ma'am the second question was given wrong. I've typed its actual question:
Q.2. An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn, if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from urn, probability that the second ball is red, is: [JEE Main 2019]
(A) 27/49
(B) 32/49 ✔
(C) 21/49
(D) 26/49
Ans.2: Let us define the events as:
E₁: Green ball is drawn first
E₂: Red ball is drawn first
A: Second ball drawn is red
Now, the probabilities are as follows:
P(E₁) = ²C₁/⁷C₁ = 2/7
P(E₂) = ⁵C₁/⁷C₁ = 5/7
P(A|E₁) = ⁶C₁/⁷C₁ = 6/7
P(A|E₂) = ⁴C₁/⁷C₁ = 4/7
By using the total probability theorem,
P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂)
= (2/7)(6/7) + (5/7)(4/7)
= 12/49 + 20/49
= 32/49
Hence, option (B) is correct.

The session was superb ma'am. Thank you so much ma'am. 😊😊😊

[me]

similar session for permutation please

[TamilIITian]

Mam am really sry for asking this !
how did u get such a jawline ?😅