1 2 3 4
h h h h
h h h t
h h t h
h h t t
h t h h
h t h t
h t t h
h t t t
t h h h
t h h t
t h t h
t h t t
t t h h
t t h t
t t t h
t t t t
1-1 1-2 1-3 1-4 1-5 1-6 2-1 2-2 2-3 2-4 2-5 2-6 3-1 3-2 3-3 3-4 3-5 3-6 4-1 4-2 4-3 4-4 4-5 4-6 5-1 5-2 5-3 5-4 5-4 5-6 6-1 6-2 6-3 6-4 6-5 6-6 So there ARE 36 possible outcomes, you see. Answer BY: Magda Krysnki (grade sevener) :P
No
Yes, providing it has 4 sides.
There are 16 possible triangles.
Sum of any two sides of a triangle is greater than third side. 3 + 4 = 7 < 8, but 7 is less than 8. So, it is not possible to form triangle with sides of length 3, 4 and 7 units.
There are 4 possible outcomes. There are 2 outcomes (heads or tails) on the first toss and 2 on the second toss. The possibilities are HH, TT, HT and TH.
Each coin has two possible outcomes, either Heads or Tails. Then the number of outcomes when all 4 coins are tossed is, 2 x 2 x 2 x 2 = 16.
2^4 = 16
4 outcomes because you have 2 coins and you multipy that by the 2 sides 2 X 2 = 4
8 outcomes are possible in this situtation. You just have to multiply 4 by 2 to get the answer.
4
4
For each of the coins, in order, you have two possible outcomes so that there are 2*2*2*2 = 16 outcomes in all.
If you toss the coins once only, it is 1/4.
2*2*2*2 = 16
If each coin is a different color, then there are 32 possible outcomes. If you can't tell the difference between the coins, and you're just counting the number of heads and tails, then there are 6 possible outcomes: 5 heads 4 heads 3 heads 2 heads 1 heads all tails
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%.