Given that <span>a bag contains 26 tiles marked with the
letters A through Z.
The probability of picking a letter from the name JACK is 4 / 26
The probability of picking a letter from the name BEN is 3 / 26.
Therefore, the probability of picking a letter from
the name JACK or from the name BEN iis given by 4 / 26 + 3 / 26 = 7 / 26 </span>
Answer:
The probability is 0.081
Step-by-step explanation:
Here, we want to calculate the probability that the 3rd inspection will be defective.
What this means is that the first two would
not be defective.
Probability of having a defective DVD = 10% = 10/100 = 0.1
Probability of not having a defective DVD = 1 -0.1 = 0.9
So the probability of third being defective = Probability of first not defective * Probability of second not defective * Probability of third defective
= 0.9 * 0.9 * 0.1 = 0.081
he magician starts with the birthday boy and moves clockwise, passing out 100100100100 pieces of paper numbered 1111 through 100100100100. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer 1111 through 100100100100, and chooses the volunteer with that number.
Method2: The magician starts with the birthday boy and moves counter-clockwise, passing out 75757575 pieces of paper numbered 1111 through 75757575. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer 1111 through 75757575, and chooses the volunteer with that number.
Method 3\: The magician starts with the birthday boy and moves clockwise, passing out 30303030 pieces of paper numbered 1111 through 30303030. He cycles around the circle until all the pieces are distributed. He gives #1111 to the birthday boy, #2222 to the next kid, and so on. He then counts the number of windows in the room and chooses the volunteer with that number.
yes probabilites can be used to make fair ones
thanx
heya
