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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

A number is drawn at random from the numbers 1 to 50. Find the probability that it is divisible by 2 or 3 or 10 - Mathematics and Statistics

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प्रश्न

A number is drawn at random from the numbers 1 to 50. Find the probability that it is divisible by 2 or 3 or 10

बेरीज

उत्तर

One number can be drawn at random from the numbers 1 to 50 in 50C1 = 50 ways.

∴ n(S) = 50

Let event A : The number drawn is divisible by 2.

∴ A = {2, 4, 6, 8, 10, …, 48, 50}

∴ n(A) = 25

∴ P(A) = `("n"("A"))/("n"("S")) = 25/50`

Let event B : The number drawn is divisible by 3.

∴ B = {3, 6, 9, 12, …, 48}

∴ n(B) = 16

∴ P(B) = `("n"("B"))/("n"("S")) = 16/50`

Let event C : The number drawn is divisible by 10.

∴ C = {10, 20, 30, 40, 50}

∴ n(C) = 5

∴ P(C) = `("n"("C"))/("n"("S")) = 5/50`

Now, A ∩ B = {6, 12, 18, 24, 30, 36, 42, 48}

∴ n(A ∩ B) = 8

∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 8/50`

B ∩ C = {30}

∴ n(B ∩ C) = 1

∴ P(B ∩ C) = `("n"("B" ∩ "C"))/("n"("S")) = 1/50`

A ∩ C = {10, 20, 30, 40, 50}

∴ n(A ∩ C) = 5

∴ P(A ∩ C ) = `("n"("A" ∩ "C"))/("n"("S")) = 5/50`

A ∩ B ∩ C = {30}

∴ n(A ∩ B ∩ C) = 1

∴ P(A ∩ B ∩ C) = `("n"("A" ∩ "B" ∩ "C"))/("n"("S")) = 1/50`

∴ P(the number is divisible by 2 or 3 or 10)

= P(A ∪ B ∪ C)

= P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

= `25/50 + 16/50 + 5/50 - 8/50 - 1/50 - 5/50 + 1/50`

= `33/50`.

shaalaa.com
Concept of Probability
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Probability - Exercise 9.2 [पृष्ठ २०१]

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