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Complete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan. Buy NowComplete Study Material based on Class 12th Syllabus, 10000+ Question Bank, Unlimited Chapter-Wise and Subject-Wise Mock Tests, Study Improvement Plan. Buy Now- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan. Buy Now- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan. Buy Now- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan. Buy NowAnswer  Hint: Here we have to find the probability that a leap year will have 53 Fridays or 53 Saturdays. We know that the leap year has 366 days and also leap year has 52 weeks and 2 extra days. We will find the sample space for these two days. Then we will count the number of possibilities that these two days are either Friday or Saturday or both. From there, we will find the required probability. Complete step-by-step answer: We know that the leap year has 366 days.We will find the number of weeks in a leap year. For that, we will divide the total days by 7.Therefore,Number of weeks \[ = \dfrac{{366}}{7} = 52\] weeks \[ + 2\] days Thus, leap year has 52 weeks and 2 extra days.We will find the sample space for these two extra days.The sample space for these two extra days is given below:\[S = \left\{ \begin{array}{l}\left( {Sunday,Monday} \right),\left( {Monday,Tuesday} \right),\left( {Tuesday,Wednesday} \right),\left( {Wednesday,Thursday} \right),\\\left( {Friday,Saturday} \right),\left( {Friday,Saturday} \right),\left( {Saturday,Sunday} \right)\end{array} \right\}\] Therefore, there are 7 total cases or possibility for these 2 extra days.Thus, \[n\left( S \right) = 7\].But according to question, we need these 2 days have either Friday or Saturday.Let \[E\] be the event that the leap year has 53 Fridays or 53 Saturdays.Therefore, from the sample space, we get\[E = \left\{ {\left( {Friday,Saturday} \right),\left( {Friday,Saturday} \right),\left( {Saturday,Sunday} \right)} \right\}\]Therefore, \[n\left( E \right) = 3\] Now, we will find the required probability using the formula \[P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}\] .Now, we will substitute the value of \[n\left( E \right)\] and \[n\left( S \right)\] in the above formula. Therefore,\[P\left( E \right) = \dfrac{3}{7}\]Hence, the required probability is \[\dfrac{3}{7}\].Thus, the correct option is B. Note: Here we have obtained the probability of a leap year having 53 Fridays or 53 Saturdays. Probability is defined as the ratio of desired or favorable outcome to the total number of outcomes. Recommended textbooks for you A First Course in Probability (10th Edition) ISBN:9780134753119 Author:Sheldon Ross Publisher:PEARSON A First Course in Probability ISBN:9780321794772 Author:Sheldon Ross Publisher:PEARSON Text Solution `(3)/(7)``(2)/(7)``(5)/(7)``(1)/(7)` Answer : A Solution : A leap year consists of 366 days, i.e., 52 full weeks and two extra days. These two extra days can be any one of the following possible outcomes, <br> (i) Monday and Tuesday <br> (ii) Tuesday and Wednesday <br> (iii) Wednesday and Thursday <br> (iv) Thursday and Friday <br> (v) Friday and Saturday <br> (vi) Saturday and Sunday <br> (vii) Sunday and Monday. <br> Let A and B be the events that a leap year contains 53 Thursday and 53 Friday, respectively. Then, <br> `P(A)=(2)/(7),P(B)=(2)/(7)` and `P(A nn B)=(1)/(7)` <br> `therefore` Required probability is given by <br> `P(A uu B) = P(A)+P(B)-P(A nn B)` <br> `=(2)/(7)+(2)/(7)-(1)/(7)=(3)/(7)` No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 1 |
What is the probability that a leap year selected at random will contain either 53 Thursday or Friday in business statistics?
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