Heads
Tails
Tails
Tails
Heads
Tails
Heads
The relative frequency of an event is calculated by dividing the number of times the event occurs by the total number of trials. In this case, the coin was flipped 20 times, and heads appeared 7 times. Therefore, the relative frequency of getting heads is ( \frac{7}{20} ), which equals 0.35 or 35%.
When you toss a 6-sided die 7 times, each toss has 6 possible outcomes. Since the tosses are independent, you can calculate the total number of outcomes by raising the number of outcomes per toss to the power of the number of tosses: (6^7). This equals 279,936 possible outcomes.
7/10
Can you PLEASE help me with this problem:Suppose a fair coin is flipped twice and the following payoffs are assigned to each of the 4 possible outcomes: H-H:win20; H-T: win 9; T-H:lose 7; T-T: lose 16what is the ecepected value of the gamble?What is the probablity of each event?Is the gamble fair?
7/10
The probability is 1. I have flipped a coin a lot more than 7 times.
7/8
7/8
The relative frequency of an event is calculated by dividing the number of times the event occurs by the total number of trials. In this case, the coin was flipped 20 times, and heads appeared 7 times. Therefore, the relative frequency of getting heads is ( \frac{7}{20} ), which equals 0.35 or 35%.
each time you flip the coin, probability to end on either side is 50% (or 0.5) (we disregard landing on the side). So, to land on the same side 7 times, it is: 0.5^7
It depends on whether you mean exactly one head or at least one head. There are 8 possible outcomes: TTT, TTH, THH, HHH, HHT, HTT, HTH, THT For at least one head, the probability is 7/8, for exactly one head, the probability is 3/8.
When you toss a 6-sided die 7 times, each toss has 6 possible outcomes. Since the tosses are independent, you can calculate the total number of outcomes by raising the number of outcomes per toss to the power of the number of tosses: (6^7). This equals 279,936 possible outcomes.
You could use it because it shows that its just 7 times 8 flipped!
7/10
That's the same as the total probability (1) minus the probability of seven heads. So: 1 - (1/2)7 = 127/128
The number is 8!/[3!(8-3)!] where n! = 1*2*3*...*n ie 8*7*6/(3*2*1) = 56
Okay, lets write out the possible outcomes when flipping a coin 3 times: HHH, HHT, HTH, THH, TTH,THT,HTT,TTT That constitures 8 scenarios in which the coin can fall over a 3 flip trial. Now, it is known that you got "at least one head" so therefore we can rule out the no head scenario (TTT) which leaves us with 7. Of those 7 times, how many times does it fall heads exactly twice? Well, we have HHT,HTH,THH. From this you can say that it there are 3 possible outcomes in which you get exactly two heads given that you get at least one head. 3/7.