Question

Sum number 20, 21 & 22.

Answer 1

Before solving the questions, let's understand what leap years and non-leap years are.

• Leap year: A year where there are a total of 366 days with 29th February included in the 366 days. A leap year occurs once every 4 years.

• Non-leap year: A year that's not a leap year is called a non-leap year. A non-leap year has a total of 365 days and 29th February isn't one of them and is not a date in the year.

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Solution for question 20:

Probability of the month of January having 5 Mondays.

(i) In a leap year. (ii) In a non-leap year.

(i) January has a total of 31 days regardless of if the year is a leap or non-leap year. Out of the 31 days, we have 4 weeks (28 days), and there are 3 more days.

We've already got 4 Mondays in four weeks, and we'll need one more Monday in one of three days. The three extra days can be of the order;

   

• [Sunday, Monday, Tuesday]

• [Monday, Tuesday, Wednesday]

• [Tuesday, Wednesday, Thursday]

• [Wednesday, Thursday, Friday]

• [Thursday, Friday, Saturday]

• [Friday Saturday, Sunday]

• [Saturday, Sunday, Monday]

Out of the 7 total outcomes, 3 of the triads have Monday in them, therefore;

$\Longrightarrow\sf P(E) = \dfrac{No: of \ favourable \ outcomes}{Total \ Number \ Of \ Outcomes}$

Where "E" is the probability of having 5 Mondays in the month of January.

$\Longrightarrow \sf P(E) = \dfrac{3}{7}$

Since there's no change in the number of total & favorable outcomes in a leap and non-leap year, both (i) and (ii) will have 3/7 as their probability.

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Solution for question 21:

Probability of the month of February having 5 Wednesdays.

(i) In a leap year. (ii) In a non-leap year.

(i) In a leap year, February has 29 days, therefore the month consists of four full weeks and one day.

Four weeks already have 4 Wednesdays in them.

And the other one day can be a Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or a Saturday.

Out of the 7 total outcomes, 1 outcome is a Wednesday, therefore;

$\Longrightarrow\sf P(E) = \dfrac{No: of \ favourable \ outcomes}{Total \ Number \ Of \ Outcomes}$

Where "E" is the probability of having 5 Wednesdays in the month of February.

$\Longrightarrow\sf P(E) = \dfrac{1}{7}$

Answer for (i): 1/7

(ii) In a non-leap year, February has 28 days, therefore the month consists of four full weeks.

Since there are no more days, it's impossible to have a 5th Wednesday in the month, making our favorable outcomes 0.

Answer for (ii): 0

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Solution for question 22:

An integer is chosen between 0 and 100, what is the probability that it's (i) divisible by 7? (ii) Not divisible by 7?

(i) Divisible by 7

Let's write down all the multiples of 7 that range between 0 and 100.

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98

Therefore we've got 14 favorable outcomes.

Total number of outcomes = 99

[Note: 0 and 100 are not included, we're looking for all numbers between them]

$\Longrightarrow\sf P(E) = \dfrac{No: of \ favourable \ outcomes}{Total \ Number \ Of \ Outcomes}$

Where "E" is the probability of the number being divisible by 7.

$\Longrightarrow\sf P(E) = \dfrac{14}{99}$

Answer for (i): 14/99

(ii) Not divisible by 7

We know that:

$\Longrightarrow \sf 1 = P(E) + P(E')$

Where;

P(E) = Probability of an event happening.

P(E') = Probability of an event not happening.

Here, the event is the number being divisible by 7.

$\Longrightarrow \sf 1 = P(E) + P(E')$

$\Longrightarrow \sf 1 = \dfrac{14}{99} + P(E')$

$\Longrightarrow \sf 1 - \dfrac{14}{99} = P(E')$

$\Longrightarrow \sf \dfrac{99 - 14}{99} = P(E')$

$\Longrightarrow \sf \dfrac{85}{99} = P(E')$

Answer for (ii) 85/99

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Final solutions:

20 (i): 3/7

20 (ii): 3/7

21 (i): 1/7

21 (ii): 0

22 (i): 14/99

22 (ii): 85/99