There are 6 batteries. Two are defective. If you pull out two batteries, what is the probability that neither are defective?
Is this how you solve it?
4/6 X 3/5
Yes, 4/6 * 3/5 = ?
Online "*" is used to indicate multiplication, because "x" might be confused with standing for an unknown.
Yes, that is correct! To solve this problem, you need to use the concept of probability. In this case, there are a total of 6 batteries, and 2 of them are defective.
To find the probability that neither of the two pulled out batteries are defective, you need to find the probability of picking a non-defective battery first, and then the probability of picking another non-defective battery after removing one.
To calculate the probability, you divide the number of favorable outcomes (picking a non-defective battery) by the number of possible outcomes (total number of batteries).
The probability of picking a non-defective battery on the first draw is 4/6, because there are 4 non-defective batteries out of 6 total batteries remaining.
After removing one battery, there are 5 batteries left, with 3 of them being non-defective. So, the probability of picking a non-defective battery on the second draw is 3/5.
To find the overall probability of neither of the two pulled-out batteries being defective, you multiply the two probabilities together:
(4/6) * (3/5) = 12/30 = 2/5
Therefore, the probability that neither of the two batteries pulled out are defective is 2/5.