SATHEE: Probability 1 Question 4

Probability 1 Question 4

4. Let $S = 1, 2, \dots, 20$ . $A$ subset $B$ of $S$ is said to be “nice”, if the sum of the elements of $B$ is 203. Then, the probability that a randomly chosen subset of $S$ is “nice”, is

(a) $\frac{6}{2^{20}}$

(b) $\frac{4}{2^{20}}$

(d) $\frac{5}{2^{20}}$

(2019 Main, 11 Jan II)

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Answer:

Correct Answer: 4. (d)

Solution:

Number of subset of $S = 2^{20}$

Sum of elements in $S$ is $1 + 2 + \dots . + 20 = \frac{20 (21)}{2} = 210$

$∵ 1 + 2 + \dots \dots + n = \frac{n (n + 1)}{2}$

Clearly, the sum of elements of a subset would be 203, if we consider it as follows

$S - 7, S - 1, 6 S - 2, 5, S - 3, 4$ $S - 1, 2, 4$

$∴$ Number of favourables cases $= 5$

Hence, required probability $= \frac{5}{2^{20}}$

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Probability 1 Question 4

4. Let S=1,2,…,20. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then, the probability that a randomly chosen subset of S is “nice”, is

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4. Let $S = 1, 2, \dots, 20$ . $A$ subset $B$ of $S$ is said to be “nice”, if the sum of the elements of $B$ is 203. Then, the probability that a randomly chosen subset of $S$ is “nice”, is