It is 0.5*0.5 = 0.25
Select 2 cards, do not put the 1st back in the deck. This is dependent probability. The outcome of drawing the 2nd card depends on the 1st card drawn. Select a card, look at it and put it back in the deck. Select a 2nd card. These are independent of each other. One does not change the probability for selecting the 2nd.
It is 1/15.
5/36 Out of the 36 possible and equally-probable outcomes (1st die's number and 2nd die's number), 5 combinations give you a sum equal 8: * 1st die gives a 2 and 2nd die gives a 6 * 1st die gives a 6 and 2nd die gives a 2 * 1st die gives a 3 and 2nd die gives a 5 * 1st die gives a 5 and 2nd die gives a 3 * 1st die gives a 4 and 2nd die gives a 4 Therefore the probability that the sum is greater than 10 equals 5/36.
Assuming the choices are made randomly and that the chosen people are not returned to the class, the probability is 77/690 = 0.1116 approx.
Out of the 36 possible and equally-probable outcomes (1st die's number and 2nd die's number), only 3 combinations give you a sum greater than 10: * 1st die gives a 5 and 2nd die gives a 6 * 1st die gives a 6 and 2nd die gives a 5 * 1st die gives a 6 and 2nd die gives a 6 Therefore the probability that the sum is greater than 10 equals 3/36, or 1/12.
Select 2 cards, do not put the 1st back in the deck. This is dependent probability. The outcome of drawing the 2nd card depends on the 1st card drawn. Select a card, look at it and put it back in the deck. Select a 2nd card. These are independent of each other. One does not change the probability for selecting the 2nd.
.61 ^ 8 = .0191707... = 1.92% And an informal proof/explanation: P(A and B) = P(A) * P(B), which means the probability of A and B = Probability of A times the probability of B. P(1st person and 2nd person) = P(1st person) * P(2nd person) P(1st person and 2nd person and 3rd person) = P(1st person) * P(2nd person) * P(3rd person) ... P(1st person) = P(2nd person) = ... = P(8th person) P(1st-8th people) = P(1st) * P(2nd) * ... * P(8th) = P(1st person) ^ 8 .61 ^ 8 = .0191707... = 1.92%
No, but it can represent the probability of such an outcome.
It is 1/15.
It is (1/4)*(1/4) = 1/16
A Naruto demon is many different animals 1st tail: Shukaku 2nd tail: nibi 3rd tail: sanbi 4th tail: yonbi 5th tail: gobi 6th tail: rokibi 7th tail: shichibi 8th tail: hachibi 9th tail: kyuubi
5/36 Out of the 36 possible and equally-probable outcomes (1st die's number and 2nd die's number), 5 combinations give you a sum equal 8: * 1st die gives a 2 and 2nd die gives a 6 * 1st die gives a 6 and 2nd die gives a 2 * 1st die gives a 3 and 2nd die gives a 5 * 1st die gives a 5 and 2nd die gives a 3 * 1st die gives a 4 and 2nd die gives a 4 Therefore the probability that the sum is greater than 10 equals 5/36.
If 10 tickets are sold, and you and a friend buy 1 ticket each. Two winners will be selected. The probability that your friend wins 1st prize and you second prize is: P(event that friend wins 1st and you 2nd) = 1/10 x 1/9 = 0.01111... ≈ 1.11 % The probability that you win 1st and your friend 2nd is the same: P(event that you win 1st and friend 2nd) = 1/10 x 1/9 = 0.01111... ≈ 1.11 % The probability that you and your friend win either way is the sum of both: P(event that you and your friend win either prize) = 0.02222... ≈ 2.22 %
Bahamas 1st Australia 2nd
Assuming the choices are made randomly and that the chosen people are not returned to the class, the probability is 77/690 = 0.1116 approx.
Out of the 36 possible and equally-probable outcomes (1st die's number and 2nd die's number), only 3 combinations give you a sum greater than 10: * 1st die gives a 5 and 2nd die gives a 6 * 1st die gives a 6 and 2nd die gives a 5 * 1st die gives a 6 and 2nd die gives a 6 Therefore the probability that the sum is greater than 10 equals 3/36, or 1/12.
you take the 2nd cord on the right and put it on the 1st on the left