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Q: What is the probability of getting at least 2 heads when flipping a coin 3 times?

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Your question is a bit difficult to understand. I will rephrase it as follows: What is the probability of getting a head if a coin is flipped once? p = 0.5 What is the probability of getting 2 heads if a coin is flipped twice = The possible events are HT, TH, HH, TT amd all are equally likely. So the probability of HH is 0.25. What is the probability of getting at least on head if the coin is flipped twice. Of the possible events listed above, HT, TH and HH would satisfy the condition of one or more heads, so the probability is 3 x 0.25 = 0.75 or 3/4. Also, since the probability of TT is 0.25, and the probability of all events must sum to 1, then we calculate the probability of one or more heads to be 1-0.25 = 0.75

As the question is "what is the probability of getting at least one head" the correct way to answer this is to ask what is the probability of not getting any heads and then subtract this from 1.The probability of not getting a head in 4 flips = 0.54 (i.e. 0.5 * 0.5 * 0.5 * 0.5) = 1/16.Therefore the probability of getting at least one head is 1 - 1/16 = 15/16.

The probability of getting at least 1 tails is (1 - probability of getting all heads) The probability of getting all heads (no tails) is ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = 1/256 = 0.00390625 so the probability of getting at least ONE tails is 1-0.30390625 = 0.99609375 = 255/256

I'm assuming you are asking what is the probability (P) of flipping a quarter.This answer really depends upon how many times up are going to flip it.If you are flipping it once, you have a 50% chance that it will land on heads and a 50% chance that it will land on tails. Either way the sum of your probabilities will add up to 1, meaning that there is a 100% chance that something will occur (see probability rules).EX: Let H= heads and let T=tails∑P= P(H)+P(T)=0.5+0.5=1However, let's say you were going to flip a coin 3 times and were wanting to know what the probability of getting at least 1 tail was. You would approach the problem this way:P( at least 1 tail)=?Next, you want to find the compliment (the opposite of what you are starting with). So the opposite of getting one tail is getting no tails. This is the same as getting all heads.P(no tails)=P(all heads)P( all heads)= P(H)3 Heads is cubed because you are flipping the coin 3= P(0.5)3 times and want all the outcomes to be heads.= 1/8By knowing that the outcome plus its compliment add up to equal 1 you get:P( all heads) + P( at least 1 tail)=1P( at least 1 tail) = 1- P( all heads)P( at least 1 tail) = 1- 1/8P( at least 1 tail) = 7/8So the probability of flipping a coin 3 times and getting a least 1 tail is 7/8. In other words, it's very likely that it will land on tails one of those three times.

The probability of getting at least one tail in a flip of six coins is the same as the probability of not getting all heads, which is 1 - (0.56), or 0.984375.

If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8

7/8

It is approx 0.1445

For 3 coin flips: 87% chance of getting heads at least once 25% chance of getting heads twice 13% chance of getting heads all three times

Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875

The probability of tossing a coin 9 times and getting at least one tail is: P(9 times, at least 1 tail) = 1 - P(9 heads) = 1 - (0.50)9 = 0.9980... ≈ 99.8%

50%

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