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That's the same as the total probability (1) minus the probability of seven heads. So:

1 - (1/2)7 = 127/128

Q: A coin is flipped seven times What is the probability that the outcome is other than all seven heads?

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This is easiest calculated by calculating the probability that NO SINGLE heads is obtained; this is of course the complement of the question. The probability of this is 1/2 x 1/2 x 1/2 ... 7 times, in other words, (1/2)7. The complement, the probability that at least one head is obtained, is then of course 1 - (1/2)7, or a bit over 99%.

The probability of one event or the other occurring is the probability of one plus the probability of the other. The probability of getting 3 heads is the probability of 3 heads (1/23) multiplied by the probability of 4 tails (1/24) multiplied by the number of possible ways this could happen. This is 7c3 or 35. Thus the probability of 3 heads is 0.2734375. The probability of 2 tails is the probability of 2 tails (1/22) multiplied by the probability of 5 heads (1/25) multiplied by the number of ways this could happen. That is 7c5 or 21. Thus the probability of 2 tails is 0.1640625 The probability of one or the other is the sum of their probabilities: 0.1640625 + 0.2734375 = 0.4375 Thus the probability of getting 3 heads or 2 tails is 0.4375.

Your question is slightly vague, so I will pose a more defined question: What is the probability of 3 coin tosses resulting in heads exactly twice? This is a pretty easy question to answer. The three possible (winning) outcomes are: 1. Heads, Heads, Tails. 2. Heads, Tails, Heads. 3. Tails, Heads, Heads. If we look at the possible combination of other (losing) outcomes, we can easily determine the probability: 4. Heads, Heads, Heads. 5. Tails, Tails, Heads. 6. Tails, Heads, Tails. 7. Heads, Tails, Tails. 8. Tails, Tails, Tails. This means that to throw heads twice in 3 flips, we have a 3 in 8 chance. This is because there are 3 winning possibilities out of a total of 8 winning and losing possibilities.

If two events are disjoint, they cannot occur at the same time. For example, if you flip a coin, you cannot get heads AND tails. Since A and B are disjoint, P(A and B) = 0 If A and B were independent, then P(A and B) = 0.4*0.5=0.2. For example, the chances you throw a dice and it lands on 1 AND the chances you flip a coin and it land on heads. These events are independent...the outcome of one event does not affect the outcome of the other.

Dependent events.

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It is neither. If you repeated sets of 8 tosses and compared the number of times you got 6 heads as opposed to other outcomes, it would comprise proper experimental probability.

The probability to get heads once is 1/2 as the coin is fair The probability to get heads twice is 1/2x1/2 The probability to get heads three times is 1/2x1/2x1/2 The probability to get tails once is 1/2 The probability to get tails 5 times is (1/2)5 So the probability to get 3 heads when the coin is tossed 8 times is (1/2)3(1/2)5=(1/2)8 = 1/256 If you read carefully you'll understand that 3 heads and 5 tails has the same probability than any other outcome = 1/256 As the coin is fair, each side has the same probability to appear So the probability to get 3 heads and 5 tails is the same as getting for instance 8 heads or 8 tails or 1 tails and 7 heads, and so on

Dependent probability is the probability of an event which changes according to the outcome of some other event.

This is easiest calculated by calculating the probability that NO SINGLE heads is obtained; this is of course the complement of the question. The probability of this is 1/2 x 1/2 x 1/2 ... 7 times, in other words, (1/2)7. The complement, the probability that at least one head is obtained, is then of course 1 - (1/2)7, or a bit over 99%.

The probability of one event or the other occurring is the probability of one plus the probability of the other. The probability of getting 3 heads is the probability of 3 heads (1/23) multiplied by the probability of 4 tails (1/24) multiplied by the number of possible ways this could happen. This is 7c3 or 35. Thus the probability of 3 heads is 0.2734375. The probability of 2 tails is the probability of 2 tails (1/22) multiplied by the probability of 5 heads (1/25) multiplied by the number of ways this could happen. That is 7c5 or 21. Thus the probability of 2 tails is 0.1640625 The probability of one or the other is the sum of their probabilities: 0.1640625 + 0.2734375 = 0.4375 Thus the probability of getting 3 heads or 2 tails is 0.4375.

Your question is slightly vague, so I will pose a more defined question: What is the probability of 3 coin tosses resulting in heads exactly twice? This is a pretty easy question to answer. The three possible (winning) outcomes are: 1. Heads, Heads, Tails. 2. Heads, Tails, Heads. 3. Tails, Heads, Heads. If we look at the possible combination of other (losing) outcomes, we can easily determine the probability: 4. Heads, Heads, Heads. 5. Tails, Tails, Heads. 6. Tails, Heads, Tails. 7. Heads, Tails, Tails. 8. Tails, Tails, Tails. This means that to throw heads twice in 3 flips, we have a 3 in 8 chance. This is because there are 3 winning possibilities out of a total of 8 winning and losing possibilities.

999,999/2,000,000 for heads, and the same for tails * * * * * The above answer implies that there is a probability of 1/1,000,000 that the coin shows neither heads nor tails. It either stands on its rim or another image appears or the disappears into some other dimension or there is some other outcome. Not impossible, perhaps, but the probability of such an event is likely to be less than 1 in a million. So for all intents and purposes, if the coin is fair, Pr(H) = Pr(T) = 0.5

If two events are disjoint, they cannot occur at the same time. For example, if you flip a coin, you cannot get heads AND tails. Since A and B are disjoint, P(A and B) = 0 If A and B were independent, then P(A and B) = 0.4*0.5=0.2. For example, the chances you throw a dice and it lands on 1 AND the chances you flip a coin and it land on heads. These events are independent...the outcome of one event does not affect the outcome of the other.

Dependent events.

Independent events.

1.525% in other words, NOT LIKELY

These are all independent events (flipping a coin will not affect the probability of drawing a Jack) so you can get the probability of all events occurring by multiplying together the probabilities of each event occurring. In other words: P (4 or 6, 2 heads, Jack) = P(4 or 6) * P(2 Heads) * P(Jack) Now we need to look at each probability separately. Remember that: Probability = Successful Outcomes / (Successful Outcomes + Unsuccessful Outcomes) In the case of rolling a die, a successful outcome (as defined in the problem) is rolling a 4 or 6. An unsuccessful outcome is everything else (1, 2, 3, or 5). Using the formula above then: Probability (4 or 6) = 2/6 = .33 Figuring out the probability of rolling two heads is slightly different because we are talking about two flips not one. In this case we have to go back to our original formula for multiple events. Probability (2 Heads) = Prob(Head) * Prob(Head) Since we know a coin-toss has a 1/2 chance of being heads or tails: Probability (2 Heads) = .5 * .5 = .25 Finally, in the case of picking up a card from a deck, a successful outcome (as defined in the problem) is picking a Jack. There are 4 Jacks in a standard deck so there are 4 possibilities of a successful outcome. There are 48 cards in a stardard deck that are not Jacks. Therefore: Probability (Jack) = 4/52 = .077 Now we can plug these values into our combination formula to get our answer. P (4 or 6, 2 heads, Jack) = P(4 or 6) * P(2 Heads) * P(Jack) P (4 or 6, 2 heads, Jack) = .33 * .25 * .077 = .00635 There is a .635% chance of rolling a 4 or 6, flipping a heads twice, AND drawing a Jack.