In a school, four out of five students have calculators.If two students are picked at random,what is the probability that a) both have a calculator b) only one has a calculator
prob(calc) = 4/5
prob(no calc) = 1/5
prob(2 of 2 having calc) = (4/5)^2 = ..
prob(1of2 have calc) = (4/5)(1/5) + (1/5)(4/5) = ...
thank you
please can you explain further
At a school with a large number of students, 4 out of every 5 students own a calculator.
a. What is the probability that a student selected at random does not own a calculator?
b. What is the probability that two students selected at random do not own a calculator?
a) Well, if four out of five students have calculators, that means there's an 80% chance that the first student picked has a calculator. And since there are only four students with calculators to choose from, the probability of the second student also having a calculator is 4/5. To find the probability of both having calculators, we multiply these probabilities together: 80% * 80% = 64%. So the probability that both students have calculators is 64% or 0.64.
b) Now, if we want to find the probability that only one student has a calculator, we can approach it this way: The first student can either have a calculator (with 80% chance) or not have a calculator (with 20% chance). If the first student has a calculator, the second student must not have a calculator (with 4/5 chance). If the first student does not have a calculator, the second student must have a calculator (with 4/5 chance). So the overall probability of only one student having a calculator is (80% * 20%) + (20% * 80%) = 32% + 32% = 64%. Therefore, the probability that only one student has a calculator is 64% or 0.64.
To solve these probability problems, we need to determine the total number of possible outcomes and the number of favorable outcomes.
a) Probability that both have a calculator:
To find the probability that both students have a calculator, we need to determine the probability of the first student having a calculator and the second student also having a calculator.
Let's assume there are 100 students in the school. Since four out of five students have calculators, then the number of students with calculators is (4/5) * 100 = 80.
The probability that the first student has a calculator is 80/100 = 4/5.
After the first student has been picked, there are now 99 students remaining, and the number of students with calculators is reduced to 79.
The probability that the second student also has a calculator is 79/99.
To find the overall probability, we multiply the probabilities of both events.
Probability (both having a calculator) = (4/5) * (79/99)
b) Probability that only one has a calculator:
To find the probability that only one student has a calculator, we need to consider two cases:
1. The first student has a calculator, but the second student does not.
2. The first student does not have a calculator, but the second student does.
Case 1:
Probability (first student has a calculator) = 4/5
After the first student has been picked, there are now 99 students remaining, and the number of students without calculators is 20 (100 total students - 80 with calculators).
Probability (second student does not have a calculator) = 20/99
Case 2:
Probability (first student does not have a calculator) = 1 - 4/5 = 1/5
After the first student has been picked, there are still 99 students remaining, but now the number of students with calculators is reduced to 80.
Probability (second student has a calculator) = 80/99
To find the overall probability, we sum the probabilities of both cases.
Probability (only one has a calculator) = (4/5 * 20/99) + (1/5 * 80/99)
You can simplify these fractions and compute the final probabilities.