What is the probability that it will point.

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A game of chance consists of spinning an arrow that comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 Q19.1

Solution:

On the face of a game, numbers 1 to 8 is shown.
Possible outcomes = 8
(i) Probability of number 8 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 1 }{ 8 }
(ii) Odd number are 1, 3, 5, 7
Probability of a number which is an odd will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 4 }{ 8 }
= \\ \frac { 1 }{ 2 }
(iii) A number greater than 2 are 3, 4, 5, 6, 7, 8 which are 6
Probability of number greater than 2 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 6 }{ 8 }
= \\ \frac { 3 }{ 4 }
(iv) A number less than 9 is 8.

Probability of a number less than 9 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
= \\ \frac { 8 }{ 8 }

Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.

Solution:

In January, there are 31 days and in an ordinary year,
there are 365 days but in a leap year, there are 366 days.
(i) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3 days.
Probability of Monday will be = \\ \frac { 3 }{ 7 }
(ii) In January of a leap year, there are 31 days i.e. 4 weeks and 3 days
Probability of Monday will be = \\ \frac { 3 }{ 7 }

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