The answer will depend on what the 8 numbers on the spinner are!
enless you include it landing on it's side the two possible outcomes for this are: Heads and Tails
There are technically 8 possible outcomes if you are talking about the side of the coin it lands on. Each coin has 2 possible outcomes (landing on heads and landing on tails). To figure out the number of outcomes for all the coins you multiply the outcomes for all of the coins together: 2 X 2 X 2= 8.
Since there are 6 possible outcomes, and you want the probability of obtaining one of the outcomes (in your case 6), the probability of it landing on a 6 is 1/6.
If the lines between the sections had no width: 20% of Landing on 1, 20% on 2, 20% on 3, 20% on 4 and 20% on 5.
It depends on the probability of landing on 3. Without further information, the question cannot be answered.
1 in 6, with a one-sided die. You have 6 possible outcomes, all have an even chance of happening. Therefore rolling a three is a 1 in 6, or a 1/6 chance.
The probability of 2 coins both landing on heads or both landing on tails is 1/2 because there are 4 possible outcomes. Head, head. Head, tails. Tails, tails. Tails, heads. Tails, heads is different from heads, tails for reasons I am unsure of.
Anything is possible!
Assuming the coins are fair, two-sided coins, and landing on their sides is not an option, there are four possible outcomes if you consider coin a having a head and coin b having a tail being a different instance from coin a being a tail and coin be having a head. Here they are; Coin A | Coin B Heads | Tails Heads | Heads Tails....| Heads Tails....| Tails
Coin - 1/2 chance for either heads or tails Die - 16 chance of landing on 1, 2, 3, 4, 5, or 6
Not possible (it is in fact possible. I had this happen yesterday.)
There are 36 possible combinations of rolling 2 die. To get an 11 or 12, the rolls would be 5, 6 or 6, 5 or 6,6 which is 3 possible correct outcomes, So, the probability of landing on an 11 or 12 is 3/36 or 1/12.