Let:
\( P(L) = 0.7 \) → Probability of lockdown
\( P(\overline{L}) = 1 - 0.7 = 0.3 \) → No lockdown
\( P(C|L) = 0.8 \) → Probability that pandemic is controlled in one month given lockdown
\( P(C|\overline{L}) = 0.3 \) → Probability that pandemic is controlled in one month given no lockdown
Using the law of total probability:
\[ P(C) = P(L) \cdot P(C|L) + P(\overline{L}) \cdot P(C|\overline{L}) \] \[ P(C) = (0.7)(0.8) + (0.3)(0.3) = 0.56 + 0.09 = 0.65 \] Correct Answer: 0.65
Let L be the event that there will be a lockdown.
Let C be the event that the pandemic is controlled in one month.
We are given the following probabilities:
We want to find the probability that the pandemic will be controlled in one month, which is \(P(C)\).
We can use the Law of Total Probability. The event C can occur in two mutually exclusive ways: either there is a lockdown and the pandemic is controlled (\(C \cap L\)), or there is no lockdown and the pandemic is controlled (\(C \cap L'\)).
\[ P(C) = P(C \cap L) + P(C \cap L') \]
Using the definition of conditional probability, \(P(C \cap L) = P(C|L)P(L)\) and \(P(C \cap L') = P(C|L')P(L')\).
So, the formula becomes:
\[ \mathbf{P(C) = P(C|L)P(L) + P(C|L')P(L')} \]
Substitute the given values into the formula:
\[ P(C) = (0.8)(0.7) + (0.3)(0.3) \]
\[ P(C) = 0.56 + 0.09 \]
\[ P(C) = \mathbf{0.65} \]
The probability that the pandemic will be controlled in one month is 0.65.
Comparing this with the given options, the correct option is:
0.65
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: