To calculate the probability of getting only one head and an even number on the tetrahedral die, we need to consider the total number of possible outcomes. There are 2 outcomes for the coin toss (HH, HT), and 2 outcomes for the tetrahedral die (2, 4). Therefore, there are a total of 2 x 2 = 4 possible outcomes. The favorable outcome for getting only one head and an even number is (HT, 2) or (HT, 4), which is 2 out of the 4 possible outcomes. Thus, the probability is 2/4 = 0.5 or 50%.
1/12
1 in 12 chance
It is (5/6)2 = 25/36
The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.
7878
50%.
1/12
1 in 12 chance
the probablity of getting a head is 1/2 and 7 heads consecutively is (1/2)^7
It is (5/6)2 = 25/36
The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.
7878
48
There are 72 permutations of two dice and one coin.
The probability of getting five heads out of 10 tosses is the same as the probablity of getting five tales out of ten tosses. One. It will happen. When this happens, you will get zero information. In other words, this is the expected result.
Tossed - Toased
The number of sequences is 27 or 128.