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An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible. P(A) = 0.22, P(B) = 0.38 - Mathematics

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प्रश्न

An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible.

P(A) = 0.22, P(B) = 0.38, P(C) = 0.16, P(D) = 0.34

बेरीज

उत्तर

When A, B, C, D are the possible exclusive and exhaustive events the P(A) + P(B) + P(C) + P(D) = 1.

P(A) = 0.22, P(B) = 0.38, P(C) = 0.16, P(D) = 0.34

Now P(A) + P(B) + P(C) + P(D) = 1

0.22 + 0.38 + 0.16 + 0.34 = 1.10 = ≠1

∴ The assignment of probability is not permissible.

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Probability
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Introduction to probability theory - Exercise 12.1 [पृष्ठ २४६]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 12 Introduction to probability theory
Exercise 12.1 | Q 1. (ii) | पृष्ठ २४६

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