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Question 36 of 43
Q.

A random variable XX denotes the number of accidents per year in a factory and the probability distribution of XX is given below :

X=xX = x01234
p(x)p(x)4K4K15K15K25K25K5K5KKK
  1. Find the constant KK.
  2. Find the probability of the event that one or two accidents will occur in this factory during the year.
Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2025Subjective· 3mImportance★★★★★
84% · 36/43 Questions
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Sum of probabilities =50K=1⇒K=0.02= 50K = 1 \Rightarrow K = 0.02; then P(X=1 or 2)=40K=0.8P(X = 1 \text{ or } 2) = 40K = 0.8.

GSEB Class-12 Statistics, Random Variable and Probability Mass Function:

Given distribution:

X=xX = x01234
p(x)p(x)4K4K15K15K25K25K5K5KKK

(i) Find KK. Since the total probability must equal 11:

4K+15K+25K+5K+K=14K + 15K + 25K + 5K + K = 1

50K=1⇒K=150=0.0250K = 1 \quad \Rightarrow \quad K = \frac{1}{50} = 0.02

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