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in Probability Distribution by (20 points)
Find the probability that the month of June may have 5 Mondays in a year. (A) \( \frac{2}{7} \) (B) \( \frac{1}{6} \) (C) \( \frac{5}{6} \) (D) \( \frac{1}{7} \)

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→ 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.

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