please help me on this;
In the mass production of bolts it is found that 5% are defective. Bolts are
selected at random and put into packets of ten. A packet is selected at
random. Find the probability that it contains:
A. no defective bolts,
B. three defective bolts, and
C. less than three defective bolts.
0.0001
To solve these types of problems, we can use the concept of probability. Probability, denoted by P, is a measure of the likelihood of an event occurring. In this case, we need to find the probability of certain events happening based on the given information.
Let's break down the problem:
A. Find the probability that the packet contains no defective bolts.
To find this probability, we need to consider that the probability of selecting a non-defective bolt is 1 - (probability of selecting a defective bolt). Since it is mentioned that 5% of the bolts are defective, the probability of selecting a defective bolt is 5% or 0.05. Therefore, the probability of selecting a non-defective bolt is 1 - 0.05 = 0.95.
Since there are ten bolts in a packet, we need to find the probability of selecting a non-defective bolt ten times in a row. This can be calculated by raising 0.95 to the power of 10: P(no defective bolts) = 0.95^10.
B. Find the probability that the packet contains three defective bolts.
To find this probability, we need to consider the number of ways we can choose three defective bolts from the ten bolts in the packet. This can be calculated using the combination formula, denoted as C(n, r), where n is the total number of items and r is the number of items we want to choose.
In this case, there are ten bolts, and we want to choose three defective bolts. Therefore, the number of ways to choose three defective bolts from ten is C(10, 3). We can calculate this as:
C(10, 3) = 10! / (3! * (10-3)!)
The probability of choosing three defective bolts from the packet is the product of the probability of selecting a defective bolt (0.05) raised to the power of three and the probability of selecting a non-defective bolt (0.95) raised to the power of seven.
C. Find the probability that the packet contains less than three defective bolts.
To find this probability, we need to calculate the probabilities of selecting zero, one, or two defective bolts and sum them up.
The probability of selecting zero defective bolts is calculated the same way as in part A, which is (0.95^10).
The probability of selecting one defective bolt is obtained by multiplying the probability of choosing a defective bolt (0.05) by the probability of choosing a non-defective bolt (0.95) raised to the power of nine (since there are nine non-defective bolts). Thus, P(selecting one defective bolt) = (0.05)*(0.95^9).
The probability of selecting two defective bolts is calculated by multiplying the probability of selecting two defective bolts (0.05^2) by the probability of selecting eight non-defective bolts (0.95^8). Thus, P(selecting two defective bolts) = (0.05^2)*(0.95^8).
To find the probability that the packet contains less than three defective bolts, we sum up the probabilities obtained from these three cases: P(packet contains less than three defective bolts) = P(selecting zero defective bolts) + P(selecting one defective bolt) + P(selecting two defective bolts).
I hope this explanation helps you understand how to solve this problem using probability concepts!