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IIT JEE Main Maths -Unit 13 -Probability- Exam Style Questions- New Syllabus

IIT JEE Main Maths - Exam Style Practice Questions - All Chapters
Question 
A coin is tossed three times. Let \( X \) denote the number of times a tail follows a head. If \( \mu \) and \( \sigma^2 \) denote the mean and variance of \( X \), then the value of \( 64(\mu + \sigma^2) \) is:
(1) 51
(2) 48
(3) 32
(4) 64
▶️ Answer/Explanation

Possible outcomes:

HHH(0), HHT(0), HTH(1), HTT(0), THH(1), THT(1), TTH(1), TTT(0).

\( P(X=0) = \frac{4}{8},\ P(X=1) = \frac{4}{8} \).

Mean: \( \mu = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2} \).

Variance: \( \sigma^2 = \frac{1}{2} – \left( \frac{1}{2} \right)^2 = \frac{1}{4} \). \[ 64(\mu + \sigma^2) = 64\left( \frac{1}{2} + \frac{1}{4} \right) = 48 \]

Hence, the correct answer is (2) 48.
Question 
Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is \( \frac{m}{n} \), where \( \gcd(m, n) = 1 \), then \( m + n \) is equal to :
(1) 14
(2) 4
(3) 11
(4) 13
▶️ Answer/Explanation

Ans. (1)

Sol. \( P = \frac{\frac{6}{10} \times \frac{5}{9}}{\left( \frac{4}{10} \times \frac{6}{9} \right) + \left( \frac{6}{10} \times \frac{5}{9} \right)} = \frac{6 \times 5}{4 \times 6 + 6 \times 5} = \frac{30}{54} = \frac{5}{9} \)

\( m = 5, n = 9 \)

\( m + n = 14 \)

Question 
If $A$ and $B$ are two events such that $P(A \mid B) = 0.1$, and $P(A \mid B)$ and $P(B \mid A)$ are the roots of the equation $12x^2 – 7x + 1 = 0$, then the value of $\frac{P(A \mid B)}{P(A \mid B)}$ is:
(1) $\frac{5}{3}$
(2) $\frac{4}{3}$
(3) $\frac{9}{4}$
(4) $\frac{7}{4}$
▶ Answer/Explanation
Ans. (3)
Sol. $12x^2 – 7x + 1 = 0$
$x = \frac{1}{3}, \frac{1}{4}$
Let $P(\frac{A}{B}) = \frac{1}{3}$ & $P(\frac{B}{A}) = \frac{1}{4}$
$P(A \cap B) = \frac{1}{3} P(B)$ & $P(A \cap B) = \frac{1}{4} P(A)$
$\Rightarrow P(B) = 0.3$ & $P(A) = 0.4$
$P(A \cup B) = P(A) + P(B) – P(A \cap B)$
= $0.3 + 0.4 – 0.1 = 0.6$
Now $\frac{P(A \mid B)}{P(A \mid B)} = \frac{P(A \cap B)}{P(A \cap B)}$
$\frac{1 – P(A \cap B)}{1 – P(A \cup B)} = \frac{1 – 0.1}{1 – 0.6} = \frac{9}{4}$
Question 

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

(1) \(\frac{1}{2}\)
(2) \(\frac{3}{5}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{9}\)

▶️ Answer/Explanation

Ans. (1)

Sol. \((a,b) = (1,3), (3,1), (2,2), (2,3), (3,2), (1,4), (4,1)\)
Required probability \(\frac{2 \cdot 2 + 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 2 + 2 \cdot 1}{6 \cdot 6} = \frac{18}{36} = \frac{1}{2}\)

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