
The probability that in a family of four children, there will be at least one boy is.
Answer
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Hint: It is better that we find the probability of all the four children being girls and subtract it from 1 to get the probability of having at least one boy. The total possible outcomes would be 16, as two options for each child (a boy or a girl) and the favourable outcome for all children being girl is just 1. Recollect the formula for finding the probability and then use the data deduced above to get the answer.
Complete step-by-step answer:
Probability in simple words is the possibility of an event to occur.
Probability can be mathematically defined as $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .
Now, let’s move to the solution to the above question. It is better that we find the probability of all the four children being girls and subtract it from 1 to get the probability of having at least one boy. As all the cases other than the case in which all are girls, there will be one boy.
Now, the total possible outcomes would be 16, as there are two options for each child, either the child is a boy or the child is a girl. Also, there are four children. So, total outcomes $={{2}^{4}}=16$ .
Now if we see the number of outcomes, in which all the children are girls is just 1, as all four children have one option, i.e., they are girls.
$\text{P}\left( \text{all girls} \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}=\dfrac{1}{16}$
Now, let us find the probability that at least one child is a boy by subtracting the above result from one.
$\text{P}\left( \text{at least one boy} \right)=1-P\left( \text{all girls} \right)=1-\dfrac{1}{16}=\dfrac{15}{16}$
Therefore, the answer to the above question is $\dfrac{15}{16}$ .
Note: In questions related to probability, be very careful about the words used in the questions, like at least, at most etc. As for the above question the probability of having at least one boy was $\dfrac{15}{16}$ , while the probability of having exactly one boy would be $\dfrac{4}{16}$ .
Complete step-by-step answer:
Probability in simple words is the possibility of an event to occur.
Probability can be mathematically defined as $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .
Now, let’s move to the solution to the above question. It is better that we find the probability of all the four children being girls and subtract it from 1 to get the probability of having at least one boy. As all the cases other than the case in which all are girls, there will be one boy.
Now, the total possible outcomes would be 16, as there are two options for each child, either the child is a boy or the child is a girl. Also, there are four children. So, total outcomes $={{2}^{4}}=16$ .
Now if we see the number of outcomes, in which all the children are girls is just 1, as all four children have one option, i.e., they are girls.
$\text{P}\left( \text{all girls} \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}=\dfrac{1}{16}$
Now, let us find the probability that at least one child is a boy by subtracting the above result from one.
$\text{P}\left( \text{at least one boy} \right)=1-P\left( \text{all girls} \right)=1-\dfrac{1}{16}=\dfrac{15}{16}$
Therefore, the answer to the above question is $\dfrac{15}{16}$ .
Note: In questions related to probability, be very careful about the words used in the questions, like at least, at most etc. As for the above question the probability of having at least one boy was $\dfrac{15}{16}$ , while the probability of having exactly one boy would be $\dfrac{4}{16}$ .
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