The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed four times, what is the probability that on at least one of the tosses the coin will turn up tails?
A. 1/16
B. 1/8
C. 1/2
D. 7/8
E. 15/16
The OA is E.
Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.
The probability is 1/2 that a certain coin will turn...
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Solving directly for the probability that we will get tails at least once is somewhat complicated - we have to think about the probability of getting tails on one, two, three, or four tosses, as well as in which order the tails will appear. However, we can solve this problem easily by instead finding the probability that we will get heads every single toss, then subtracting from one. This gives us the probability that we will NOT get heads every single toss - in other words, we will get tails at least once.
The probability we will get heads on each toss is 1/2. Multiplying by 1/2 for each toss gives
$$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{16}$$
So the probability that we will get heads every single toss is 1/16.
Subtracting this from one gives
$$1-\frac{1}{16}=\frac{15}{16}$$
So the probability that we will get tails at least once is 15/16, or answer choice E.
When solving for the probability that something will occur, it's often easiest to subtract out the cases that won't occur. Not always the case, but useful to keep in your tool belt!
The probability we will get heads on each toss is 1/2. Multiplying by 1/2 for each toss gives
$$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{16}$$
So the probability that we will get heads every single toss is 1/16.
Subtracting this from one gives
$$1-\frac{1}{16}=\frac{15}{16}$$
So the probability that we will get tails at least once is 15/16, or answer choice E.
When solving for the probability that something will occur, it's often easiest to subtract out the cases that won't occur. Not always the case, but useful to keep in your tool belt!
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When it comes to probability questions involving "at least," it's best to try using the complement.swerve wrote:The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed four times, what is the probability that on at least one of the tosses the coin will turn up tails?
A. 1/16
B. 1/8
C. 1/2
D. 7/8
E. 15/16
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 tails) = 1 - P(NOT getting at least 1 tails)
What does it mean to NOT get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)
Now let's calculate P(getting zero tails)
What needs to happen in order to get zero tails?
Well, we need heads on the 1st toss and heads on the 2nd toss and heads on the 3rd toss and heads on the 4th toss
We can write P(getting zero tails) = P(heads on 1st AND heads on 2nd AND heads on 3rd AND heads on 4th)
This means that P(getting zero tails) = P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd) x P(heads on 4th)
Which means P(getting zero tails) = (1/2)x(1/2)x(1/2) x(1/2) = 1/16
We're now ready to answer the question.
P(getting at least 1 tails) = 1 - P(NOT getting at least 1 tails)
= 1 - 1/16
= 15/16
= E
Cheers,
Brent
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The only way no tails will turn up is all four tosses turn up heads, which has a probability of 1/2 x 1/2 x 1/2 x 1/2 = 1/16. Therefore, the probability that at least one toss will turn up tails is 1 - 1/16 = 15/16.swerve wrote:The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed four times, what is the probability that on at least one of the tosses the coin will turn up tails?
A. 1/16
B. 1/8
C. 1/2
D. 7/8
E. 15/16
.
Answer: E
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