If a coin is tossed twice, find the probability of getting at least one head.
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{3}{4}$
D. $\dfrac{2}{3}$
Answer
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Hint: Here, a coin is tossed two times and we are given the condition – at least one head appears i.e., we have to find the probability of at least one head as well as the probability of getting more than one head. By using the formula Probability = favourable outcomes / total possible outcomes we will find the required solution.
Complete step by step Answer:
Let a coin be tossed two times. Let H denote the coming of a Head and T represents the appearance of the Tail of the coin which is being tossed.
If a coin is tossed twice, the possible outcomes will be HH, H T, T H, T T.
By the definition of the probability,
Probability = favourable outcomes / total possible outcomes
In this case, favourable outcomes will be those which will contain at least one head H or more than one head H.
Therefore, the possible outcomes of getting at least one head (one head or more than one head) are HH, HT, T H. they sum up to a total of 3 outcomes of getting at least one head.
$\therefore $ probability of getting at least one head = $\dfrac{3}{4}$
Therefore, option (c) $\dfrac{3}{4}$ is correct.
Note: Whenever there’s a situation where “at least” is mentioned, the possible outcomes will always include the possibility of getting the mentioned times as well as more than the given condition.
Probability is defined as the branch of mathematical statistics which tells about how likely an event appears to occur. The probability of a number always lies between 0 and 1.
In the above question, $\dfrac{3}{4}$= 0.75 which lies between 0 and 1.
Complete step by step Answer:
Let a coin be tossed two times. Let H denote the coming of a Head and T represents the appearance of the Tail of the coin which is being tossed.
If a coin is tossed twice, the possible outcomes will be HH, H T, T H, T T.
By the definition of the probability,
Probability = favourable outcomes / total possible outcomes
In this case, favourable outcomes will be those which will contain at least one head H or more than one head H.
Therefore, the possible outcomes of getting at least one head (one head or more than one head) are HH, HT, T H. they sum up to a total of 3 outcomes of getting at least one head.
$\therefore $ probability of getting at least one head = $\dfrac{3}{4}$
Therefore, option (c) $\dfrac{3}{4}$ is correct.
Note: Whenever there’s a situation where “at least” is mentioned, the possible outcomes will always include the possibility of getting the mentioned times as well as more than the given condition.
Probability is defined as the branch of mathematical statistics which tells about how likely an event appears to occur. The probability of a number always lies between 0 and 1.
In the above question, $\dfrac{3}{4}$= 0.75 which lies between 0 and 1.
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