If you are drawing two cards from a full deck of cards (without jokers) then the probability will depend upon whether the the first card is replaced before the second is drawn, but the probability will also be different to being dealt a hand whilst playing Bridge (or Whist), which will again be different to being dealt a hand at Canasta.
Without the SPECIFIC context of the two cards being got, I cannot give you a more specific answer.
The probability of rolling a 7 with 2 dice is 6/36; probability of rolling an 11 is 2/36. Add the two together to find probability of rolling a 7 or 11 which is 8/36 or 2/9.
7/8
7
The probability of getting on the first draw a black ball is: P(B1) = 3/7. The probability of getting a red ball given the event of drawing a black ball on the first draw is: P(R2│B1) = 4/6. The probability of drawing a black ball on the first draw and a red ball on the second draw is: P(B1UR2) = P(B1)∙P(R 2│B1) = (3/7)∙(4/6) = 0.2857... ~ 0.286 ~ ~ 28.6%
The probability of drawing 3 red marbles in a row from a bag of 3 blue and 4 red marbles without replacement is right at about 0.11 or about 11 in 100, if you want odds.Remember that probability is a pure number between zero (no chance - it's impossible) and 1 (it will happen - it's a certainty). A probability of 0.5 is a 50-50 shot, like a coin toss. We good? Let's move on.We have 7 marbles, and there is a 4 in 7 shot of getting a red marble in the first draw. There is a 3 in 6 shot the next draw since now we only have 6 marbles total (NO replacement) and only 3 red ones (we already drew a red one out). Similarly, there is a 2 in 5 shot of getting a third red one if the first two are red. The probability of the first action (getting a red marble on the first draw) is 4/7 or a probability of about 0.5714. Second draw, it's 3/6 or a probability of 0.5. Last, it's 2/5 or a probability of 0.40. To find the intersection of these probabilities, we just multiply them all together, and we'll get right at about 0.011 for a probability.
The probability of getting a particular suit (hearts, spades, diamonds, clubs) is 1 in 4. The probability of getting a card less than 8 (2, 3, 4, 5, 6, 7) is 6 in 13. The probability, then, of getting a particular suit less than 8 is (1 in 4) times (6 in 13) or 6 in 52 or 3 in 26.
Probability that the sum is 6 = 5/36 Probability that the sum is 7 = 6/36
The probability of getting a 7 on one roll of a die is zero.If you meant to ask about two dice, the probability is 6 in 36, or 1 in 6.
First we find the probability of getting a 7. Of the 36 outcomes possible 6 result in a sum of 7, in other words 1/6. The probability of getting an 11 is 2/36 or 1/18. The probability of getting one or the other is the sum of the two, 8/36 or 2/9. The proability of getting neither is equal to the probability of getting anything other than 7 or 8. We find this value by subtracting 2/9 from 1. So the probability of not getting 7 or 11 is 7/9.
The probability is 1 as all values on a die are less than 7.
The probability of rolling a 7 at any time on a single die is zero.
The probability of getting only one tails is (1/2)7. With seven permutations of which flip is the tails, this gives a probability of: P(six heads in seven flips) = 7*(1/2)7 = 7/128
The probability is 1/42.
Find the probability of getting a sum of 7 and call that p(7). Now the complement rule says 1-p(7) is the probability of not getting 7.The ways we get 7 are 1 and 6, or 2 and 5, or 3 and 4.If you make a table with 1-6 on the top and 1-6 on the side, you can let each entry in your table be the sum of the top row and the side column. Since this is a 6x6 table, you can seethere are 36 outcomes.6 of those outcome result in a sum of 7.They are:1+66+12+55+23+44+3So p(7) =6/36 or 1/6. P of not getting 7 is 1-1/6=5/6
The experiment is not defined! The probability of the event described, when 9 or more number cubes are rolled, is 0.
7
As the maximum value of the dots on the face of a traditional dice is 6 the probability of throwing A die with the value of less than 7 is 100%.