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Bunuel
What is the probability of landing on tails at least once in a sequence of 100 tosses of a specific coin?

To answer the question, we need to know two things: the possible outcomes and the probability of each outcome.

(1) When tossing the coin, there are only two possible outcomes, heads or tails, and both have an equal chance of occurring.

This implies that there are two possible outcomes, heads or tails, and the probability of each is 1/2. Sufficient.

(2) In a series of 100 tosses of the coin, we can expect almost 50 heads.­

This statement is very imprecise and does not provide information about the possible outcomes or the probability of each outcome. Not sufficient.

Answer: A.
Bunuel what are you referrng to as outcome here?

Does it mean P( 1tail 99 heads) is 1 outcome or does it mean P(1 Tail on one toss) is 1 outcome
­
In this context, "outcome" refers to the number of possible results from the coin tosses.­
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if it is almost 50 (b) , then the probality of at least 1 tail is 100%??
why not of only 50 times head come bb
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anish0953
if it is almost 50 (b) , then the probality of at least 1 tail is 100%??
why not of only 50 times head come bb
­
bb is a whole different person. As for your question, "almost 50" isn't exact. Is it 49? 48? ... Each of these gives a different answer to the question.
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silly tricky question. statement 2 looks like a 50/50 probability to me.
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We know that in real life every coin has 50/50 chance of getting H or T
but it doesn't necessarily happen that we will always get 5 heads in 10 tosses
It might also happen that we might get 0 heads

So why have we assumed that there will definitely be 1 tail?

@ [color=#683d3d]KarishmaB[/color] Bunuel
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ADisHere
We know that in real life every coin has 50/50 chance of getting H or T
but it doesn't necessarily happen that we will always get 5 heads in 10 tosses
It might also happen that we might get 0 heads

So why have we assumed that there will definitely be 1 tail?

@ [color=#683d3d]KarishmaB[/color] Bunuel
The solution does not assume there will definitely be 1 tail. It calculates the probability of at least one tail.
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I do not like statement 2. "Almost" means less than 50. We expect 47, 48, or 49; but not 50 or more. But in this case we can expect 50 or more. Semantically, statement 2 is false, and thus the problem is wrong.
guddo
­What is the probability of landing on tails at least once in a sequence of 100 tosses of a specific coin?

(1) When tossing the coin, there are only two possible outcomes, heads or tails, and both have an equal chance of occurring.

(2) In a series of 100 tosses of the coin, we can expect almost 50 heads.­


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Hello, @Bunuel

Hope you are doing well. I wanted your support in understanding that why cannot we say that the possible number of outcomes would be 2^100? Because the coin is tossed 100 times. Also, why cannot we say that the probability of P(H) and P(T) is 1/2 each? When I read the question, I put down these two information, like for me, it was already given. Did I assume information about 2^100 and P(H) and P(T) is 1/2 each?

Thank you for your support.

Best,
Komal

Bunuel
What is the probability of landing on tails at least once in a sequence of 100 tosses of a specific coin?

To answer the question, we need to know two things: the possible outcomes and the probability of each outcome.

(1) When tossing the coin, there are only two possible outcomes, heads or tails, and both have an equal chance of occurring.

This implies that there are two possible outcomes, heads or tails, and the probability of each is 1/2. Sufficient.

(2) In a series of 100 tosses of the coin, we can expect almost 50 heads.­

This statement is very imprecise and does not provide information about the possible outcomes or the probability of each outcome. Not sufficient.

Answer: A.
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Komal324
Hello, @Bunuel

Hope you are doing well. I wanted your support in understanding that why cannot we say that the possible number of outcomes would be 2^100? Because the coin is tossed 100 times. Also, why cannot we say that the probability of P(H) and P(T) is 1/2 each? When I read the question, I put down these two information, like for me, it was already given. Did I assume information about 2^100 and P(H) and P(T) is 1/2 each?

Thank you for your support.

Best,
Komal

Bunuel
What is the probability of landing on tails at least once in a sequence of 100 tosses of a specific coin?

To answer the question, we need to know two things: the possible outcomes and the probability of each outcome.

(1) When tossing the coin, there are only two possible outcomes, heads or tails, and both have an equal chance of occurring.

This implies that there are two possible outcomes, heads or tails, and the probability of each is 1/2. Sufficient.

(2) In a series of 100 tosses of the coin, we can expect almost 50 heads.­

This statement is very imprecise and does not provide information about the possible outcomes or the probability of each outcome. Not sufficient.

Answer: A.

You assumed information that was not explicitly given. The question asks about a specific coin, but unless it is stated that the coin is fair, you cannot assume P(H) = 1/2 and P(T) = 1/2.
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Dear Bunuel,

Understood. But, we can calculate the number of possible outcomes as 2^100 right? Because we are given that the coin is tossed 100 times.

Thank you!

Best,
Komal

Bunuel
Komal324
Hello, @Bunuel

Hope you are doing well. I wanted your support in understanding that why cannot we say that the possible number of outcomes would be 2^100? Because the coin is tossed 100 times. Also, why cannot we say that the probability of P(H) and P(T) is 1/2 each? When I read the question, I put down these two information, like for me, it was already given. Did I assume information about 2^100 and P(H) and P(T) is 1/2 each?

Thank you for your support.

Best,
Komal

Bunuel
What is the probability of landing on tails at least once in a sequence of 100 tosses of a specific coin?

To answer the question, we need to know two things: the possible outcomes and the probability of each outcome.

(1) When tossing the coin, there are only two possible outcomes, heads or tails, and both have an equal chance of occurring.

This implies that there are two possible outcomes, heads or tails, and the probability of each is 1/2. Sufficient.

(2) In a series of 100 tosses of the coin, we can expect almost 50 heads.­

This statement is very imprecise and does not provide information about the possible outcomes or the probability of each outcome. Not sufficient.

Answer: A.

You assumed information that was not explicitly given. The question asks about a specific coin, but unless it is stated that the coin is fair, you cannot assume P(H) = 1/2 and P(T) = 1/2.
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Komal324
Dear Bunuel,

Understood. But, we can calculate the number of possible outcomes as 2^100 right? Because we are given that the coin is tossed 100 times.

Thank you!

Best,
Komal


Technically, we don't even know from the question that the coin has exactly two outcomes. Maybe it's a coin that can land on its edge, or it's some kind of oddly shaped object with more than two possible outcomes. Unless the question explicitly says it's a fair two-sided coin, we can’t assume both the number of outcomes (2^100) or that P(H) = P(T) = 1/2.
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