
If three coins are tossed simultaneously, then find the probability of getting at least two heads.
A. \[\dfrac{1}{2}\]
B. \[\dfrac{3}{8}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{2}{3}\]
Answer
616.5k+ views
Hint: To solve such types of questions we first of all try to get the total number of outcomes and also the number of favorable outcomes. This helps to find the probability which is given by the number of favorable outcomes divided by the total number of outcomes.
Complete step-by-step answer:
Given three coins are tossed simultaneously, then we have to find out the probability of getting at least two heads.
We know that a coin has two sides: head (H) and tail (T).
If we toss three coins then the possible outcomes are as follows:
(H,H,H) (H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H) & (T,T,T)
\[\Rightarrow \]Number of the total possible outcomes=8.
Now the number of favorable outcomes is getting at least two heads, that can be obtained from observing total number of outcomes and are listed as,
{(H,H,H) (H,H,T),(H,T,H),(T,H,H)}.
Therefore, the number of favorable outcomes are 4.
Then applying the formula of probability as number of favorable outcomes divided by the total number of outcomes, we have,
Probability (at least 2 heads)= \[\dfrac{Number\text{ }of\text{ }favorable\text{ }outcomes}{total\text{ }number\text{ }of\text{ }outcomes}\]
\[\Rightarrow \]Probability P (at least 2 heads)= \[\dfrac{4}{8}=\dfrac{1}{2}\]
\[\Rightarrow \] Probability P (at least 2 heads)=\[\dfrac{1}{2}\]
Hence, the probability of getting at least 2 heads=\[\dfrac{1}{2}\], which is the required answer to the given question.
Note: The possibility of error in this question can be not generalizing the total possible outcomes and the possible favorable outcomes. For calculating the probability be always try to find the total outcomes before finding the number of it.
Complete step-by-step answer:
Given three coins are tossed simultaneously, then we have to find out the probability of getting at least two heads.
We know that a coin has two sides: head (H) and tail (T).
If we toss three coins then the possible outcomes are as follows:
(H,H,H) (H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H) & (T,T,T)
\[\Rightarrow \]Number of the total possible outcomes=8.
Now the number of favorable outcomes is getting at least two heads, that can be obtained from observing total number of outcomes and are listed as,
{(H,H,H) (H,H,T),(H,T,H),(T,H,H)}.
Therefore, the number of favorable outcomes are 4.
Then applying the formula of probability as number of favorable outcomes divided by the total number of outcomes, we have,
Probability (at least 2 heads)= \[\dfrac{Number\text{ }of\text{ }favorable\text{ }outcomes}{total\text{ }number\text{ }of\text{ }outcomes}\]
\[\Rightarrow \]Probability P (at least 2 heads)= \[\dfrac{4}{8}=\dfrac{1}{2}\]
\[\Rightarrow \] Probability P (at least 2 heads)=\[\dfrac{1}{2}\]
Hence, the probability of getting at least 2 heads=\[\dfrac{1}{2}\], which is the required answer to the given question.
Note: The possibility of error in this question can be not generalizing the total possible outcomes and the possible favorable outcomes. For calculating the probability be always try to find the total outcomes before finding the number of it.
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