We are given a bag containing 5 red, 7 green, and 3 blue balls. The total number of balls in the bag is:
\[ 5 + 7 + 3 = 15 \]
We are tasked with finding the probability that the second drawn ball is red. The key here is to consider the possible cases that can occur when drawing two balls sequentially.
There are two primary cases to analyze: the case where the first ball drawn is red and the case where the first ball drawn is not red.
Case 1: First ball drawn is red.
If the first ball drawn is red, there are now 4 red balls remaining out of the 14 balls left. The probability of drawing a red ball in this case is:
\[ P({second red | first red}) = \frac{4}{14} = \frac{2}{7} \] Case 2: First ball drawn is not red.
If the first ball drawn is not red (i.e., it is either green or blue), there are still 5 red balls remaining out of the 14 balls left. The probability of drawing a red ball in this case is:
\[ P({second red | first not red}) = \frac{5}{14} \] Total probability:
The total probability is found by multiplying the probability of each case by the probability of the first ball being red or not red. The probability of drawing a red ball first is \( \frac{5}{15} \), and the probability of drawing a non-red ball first is \( \frac{10}{15} \). Therefore, the total probability of the second drawn ball being red is:
\[ P({second red}) = \left( \frac{5}{15} \times \frac{2}{7} \right) + \left( \frac{10}{15} \times \frac{5}{14} \right) = \frac{1}{3} \] Thus, the probability that the second ball drawn is red is \( \frac{1}{3} \), corresponding to option (B).
P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?
If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \(\frac{P(A \cup B)}{P(A \cap B)}\)
The table shows the data of running a machine for five years. The original machine cost is Rupees 70,000. In order to minimize the average total cost per year for running the machine, the machine should be replaced after ............. years. (Answer in integer)
A company purchases items in bulk for getting quantity discounts in the item’s price. The price break-up is given in the table. The annual demand for the item is 5000 units. The ordering cost is Rupees 400 per order. The annual inventory carrying cost is 30 percent of the purchase price per unit. The optimal order size (in units) is .......... (Answer in integer)