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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