
The probability that a leap year selected at random contains 53 Sunday is
\[\begin{align}
& A.\dfrac{7}{266} \\
& B.\dfrac{26}{183} \\
& C.\dfrac{1}{7} \\
& D.\dfrac{2}{7} \\
\end{align}\]
Answer
552.9k+ views
Hint: In this question, we need to find the probability of selecting 53 Sundays in a leap year. For this, we will first find the possibilities of days that we get. After that, we will find the possibilities with Sunday, these possibilities will become our total outcomes. Using the formula of probability i.e. $\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{total outcomes}}$ we will find required probability.
Complete step by step answer:
Here we are given a leap year, we need to find the probabilities of selecting a leap year that has 53 Sundays. Let us first find the total outcomes.
We know that in a leap year we have 2 days extra after counting 52 weeks. Now, these 2 days can be any two consecutive days of a week. So possibilities can be: (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday).
As we can see there could be a total of 7 possibilities. Therefore, total outcomes = 7.
Now let us find the number of favorable outcomes.
We need to find leap year with 53 Sundays i.e. we need to find possibilities of getting an extra Sunday except for 52 already Sunday's present. Out of the 7 count possibilities, only two possibilities have Sunday which is (Saturday, Sunday) and (Sunday, Monday). Therefore, a number of favorable outcomes become equal to 2.
We know that, probability of an event is given by,
$\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{total outcomes}}$.
Putting in the values we gets,
$\text{Probability}=\dfrac{\text{2}}{\text{7}}$.
Hence required probability is equal to $\dfrac{\text{2}}{\text{7}}$.
Note:
Students should note that a normal year has one day extra after 52 weeks. So, a leap year has two days extra. Always try to write all the possibilities for proper calculation. Students often miss the pair of (Sunday, Monday). Note that, the probability is always less than 1 and greater than 0.
Complete step by step answer:
Here we are given a leap year, we need to find the probabilities of selecting a leap year that has 53 Sundays. Let us first find the total outcomes.
We know that in a leap year we have 2 days extra after counting 52 weeks. Now, these 2 days can be any two consecutive days of a week. So possibilities can be: (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday).
As we can see there could be a total of 7 possibilities. Therefore, total outcomes = 7.
Now let us find the number of favorable outcomes.
We need to find leap year with 53 Sundays i.e. we need to find possibilities of getting an extra Sunday except for 52 already Sunday's present. Out of the 7 count possibilities, only two possibilities have Sunday which is (Saturday, Sunday) and (Sunday, Monday). Therefore, a number of favorable outcomes become equal to 2.
We know that, probability of an event is given by,
$\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{total outcomes}}$.
Putting in the values we gets,
$\text{Probability}=\dfrac{\text{2}}{\text{7}}$.
Hence required probability is equal to $\dfrac{\text{2}}{\text{7}}$.
Note:
Students should note that a normal year has one day extra after 52 weeks. So, a leap year has two days extra. Always try to write all the possibilities for proper calculation. Students often miss the pair of (Sunday, Monday). Note that, the probability is always less than 1 and greater than 0.
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