Question

the probability that the month of june may have 5 monday in a non leap year​

Answer 1

Probability of getting 5 Mondays in month of June is 2/7.

Step-by-step explanation:

• Total number of days in June = 30
• Therefore, there must be 4 Mondays, 4 Tuesdays, 4 Wednesdays, ….. 4 Sundays.
• Rest days = 30 - 7 x 4 = 30 - 28 = 2.
• These 2 days may be (i) Sunday & Monday, (ii) Monday & Tuesday, (iii) Tuesday & Wednesday, (iv) Wednesday & Thursday, (v) Thursday & Friday, (vi) Friday & Saturday, (vii) Saturday & Sunday.
• Hence, total outcomes for these 2 days. (Rest 2 days) is n(S) = 7
• Outcomes favourable to 5 Mondays are Monday & Tuesday and Sunday & Monday.
• Number of outcomes favourable to get 5 Mondays is n(E) = 2.
• ∴ Probability of getting 5 Mondays in month of June is P(E) = n(E)/n(S) = 2/7.

Answer 2

The probability of that month June have 5 mondays in a non leap year is 2/7

Given :

The month of June in a non leap year

To find :

The probability of that month June have 5 mondays in a non leap year

Solution :

Step 1 of 3 :

Find total number of possible outcomes

The month of June has 30 days

30 days = 4 weeks 2 days

Now 4 weeks contains 4 Mondays

2 days is one of the below

( Sunday, Monday), ( Monday, Tuesday), (Tuesday, Wednesday), ( Wednesday, Thursday), ( Thursday, Friday), ( Friday, Saturday ), (Saturday, Sunday)

So the total number of possible outcomes = 7

Step 2 of 3 :

Find total number of possible outcomes for the event

Let A be the event that month June have 5 mondays in a non leap year

So the total event points for the event A is ( Sunday, Monday), ( Monday, Tuesday)

So the total number of possible outcomes for the event A is 2

Step 3 of 3 :

Find the probability

Hence the required probability

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{2}{7} }$

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