You have not given any information about the shape and numbering of the dice nor how many are thrown each time. Without that information it is not possible to answer the question.
1 in 2 to get tails on first toss 1 in 2 to get heads on second toss To get tails on first toss and heads on second toss, probability is 1/2 x 1/2 = 1 in 4
Both are equivalent! * Probability of using 50-50 and getting the right answer: 1/2 * Probability of using double dip and getting the right answer: = Probability that either first or the second guess is correct P(First guess is correct) = 1/4 + P(First guess in wrong and second guess is correct) = 3/4 * 1/3 = 1/4 = 1/2
Assuming it is a standard poker deck with 52 cards and 4 aces The probability of getting your first card an ace is = 4/52 Over here you need to reread your questions. There are something you need to know before i continue - when you draw your second card, did you return the first card back into the pile? lets say if you draw an ace of spades, would you be able to redraw it again for the second card? If the first ace is return to the pile: probability of getting a second ace is also 4/52 so the total probability of getting both cards an ace is (4/52 x 4/52) If the first ace is not return to the pile: probability of getting a second ace is now 3/51 note that removing one ace also removes one card from the pile. total card is now 51 with 3 aces only so the total probability of getting both cards an ace is (4/52 x 3/51) hope i help.
If you toss a coin often enough the probability is 1. The probability of 9 H in the first 9 tosses is (1/2)9 = 1/512
2/12 or 1/6
13/204
The probability is 1/2 because the second outcome has no affect on the first outcome.
The probability of throwing just one one with two dice can be calculated by considering the different ways it can occur. There are two ways to get one one: rolling a one on the first die and any number on the second die, or rolling any number on the first die and a one on the second die. There are a total of 36 possible outcomes when rolling two dice, so the probability is 2/36, which simplifies to 1/18.
This is an example of an INDEPENDENT event: The probability of throwing a double six with two dice is the result of throwing six with the first die and six with the second die. The total possibilities are, one from six outcomes for the first event and one from six outcomes for the second, Therefore (1/6) * (1/6) = 1/36th or 2.77%. The two events are independent, since whatever happens to the first die cannot affect the throw of the second, the probabilities are therefore multiplied, and remain 1/36th.
1 in 2 to get tails on first toss 1 in 2 to get heads on second toss To get tails on first toss and heads on second toss, probability is 1/2 x 1/2 = 1 in 4
It is 1/18.
The first dice can show any number. However the second dice has a 1 in 6 chance of being the same as the first. Hence the probability of getting two numbers the same is 1/6.
What is the probability of rolling a 6 the first time and a 1 the second time
Both are equivalent! * Probability of using 50-50 and getting the right answer: 1/2 * Probability of using double dip and getting the right answer: = Probability that either first or the second guess is correct P(First guess is correct) = 1/4 + P(First guess in wrong and second guess is correct) = 3/4 * 1/3 = 1/4 = 1/2
Assuming it is a standard poker deck with 52 cards and 4 aces The probability of getting your first card an ace is = 4/52 Over here you need to reread your questions. There are something you need to know before i continue - when you draw your second card, did you return the first card back into the pile? lets say if you draw an ace of spades, would you be able to redraw it again for the second card? If the first ace is return to the pile: probability of getting a second ace is also 4/52 so the total probability of getting both cards an ace is (4/52 x 4/52) If the first ace is not return to the pile: probability of getting a second ace is now 3/51 note that removing one ace also removes one card from the pile. total card is now 51 with 3 aces only so the total probability of getting both cards an ace is (4/52 x 3/51) hope i help.
If you toss a coin often enough the probability is 1. The probability of 9 H in the first 9 tosses is (1/2)9 = 1/512
2/12 or 1/6