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  • Explain Solution R.D.Sharma Class 12 Chapter 30 Probability Exercise 30.4 Question 26 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 30 Probability  Exercise 30.4 Question 26 Maths Textbook Solution.

Answers (1)

Answer: \frac{126}{1000}

Given: 

             Probability\: of\: finding\: a\: green\: signal\: on\: X=30%

Hint: Probability=\frac{No\: of\: outcomes}{Total\: outcomes}

Solution:

Given,

           Probability\: of\: finding\: a\: green\: signal\: on\: X=\frac{30}{100}=\frac{3}{10}

Probability\: of\:not\: finding\: a\: green\: signal\: on\: X=1-\frac{30}{100}=\frac{7}{10}

                Now, Let D_{1},D_{2},D_{3} be the three days for which the signal is green and D{}'_{1},D{}'_{2},D{}'_{3}   be the three days for not finding a green signal.

                We have to find probability of finding green signal on consecutive two days out of three

                So the event would be \left ( D_{1},D_{2},D{}'_{3} \right ) or\left ( D{}'_{1},D_{2},D_{3} \right )

So the probability of finding green signal on consecutive two days out of three

\begin{aligned} &=\mathrm{P}\left(D_{1}, D_{2}, D_{3}^{\prime}\right)+P\left(D_{1}^{\prime}, D_{2}, D_{3}\right) \\ &=\frac{3}{10} \times \frac{3}{10} \times \frac{7}{10}+\frac{7}{10} \times \frac{3}{10} \times \frac{3}{10} \\ &=\frac{63}{1000}+\frac{63}{1000} \\ &=\frac{126}{1000} \end{aligned}

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