probability

Question

Three numbers are chosen at random without replacement from $\left\{1, 2, 3,....., 10 \right\}$. What is the probability that minimum of the chosen numbers is $3$ or their maximum is $7$?

1. $\frac{11}{30}$
2. $\frac{11}{40}$
3. $\frac{1}{7}$
4. $\frac{1}{8}$

Comments

Answer is (B)  11/40.

Answer 1

Ans is (B) 11/40

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 it can be solved  by inclusion exclusion principle.

P(A ⋃ B)  = P(A) + P(B) - P(A ⋂ B)

Suppose, P(A) = probability that minimum is 3 among chosen 3 numbers. 

        = ⁷C₂/¹⁰C₃

_​For this 3 is already taken with no choice and for remaining ⁷C₂ is total number of ways to choose 2 numbers more than 3 (i.e. 4,5,6,7,...10) from given set.

P(B) = probability that maximum numbers is 7 among chosen 3 numbers

        = ⁶C₂/¹⁰C₃

_​For this 7 is already taken with no choice and for remaining ⁶C₂ is total number of ways to choose 2 numbers less than 7 (i.e. 0,1,2,.....6) from given set.

P(A ⋂ B) = probability that minimum number is 3 and maximum is 7 among chosen 3.

               = ³C₁/¹⁰C₃

_​For this 3 & 7 is already taken with no choice and for remaining ³C₁ is total number of ways to choose 1 number more than 3 and less than 7  (i.e. 4,5,6) from given set.

Now the probability that minimum is 3 or maximum is 7 among chosen 3 numbers i.e. P(A ⋃ B) .

P(A ⋃ B)  = P(A) + P(B) - P(A ⋂ B)

                = ⁷C₂/¹⁰C_{3 +} ⁶C₂/¹⁰C_{3 -} ³C₁/¹⁰C₃

                   = (⁷C_2 + ⁶C_2 - ³C₁)/¹⁰C₃

                = (21 + 15 - 3) / 120

                = 33/120

                = 11/40

Answer 2

favourable events from which we need to choose two more when 3 minimum is already chosen = {4, 5, 6, 7, 8, 9, 10}

similarly, favourable events from which we need to choose two more when 7 maximum is already chosen = {1, 2, 3, 4, 5, 6}

$$\begin{align*} \text{required probability} &= \frac{\binom{7}{2} + \binom{6}{2} - \binom{3}{1}}{\binom{10}{3}}\\ &= \frac{21+15-3}{120}\\ &= \frac{33}{120}\\ &= \frac{11}{40} \end{align*}$$

Comments

how ?

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Answer is (B)  11/40.

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Corrected.