Question

A month is selected at random in a year. The probability that it is March or October, is
(a)$\frac{1}{12}$
(b)$\frac{1}{6}$
(c)$\frac{3}{4}$
(d) None of these

Answer 1

SOLUTION :  

The correct option is (b) : 1/6

Given : A month is selected at random in a year .

Total months in a year = 12  

Total number of outcomes = 12

Let E = Event of selecting that the month is March or October

Favourable outcomes =  March or October  

Number of favourable outcomes to E = 2

Probability P(E) = Number of favourable outcomes / total number of outcomes

P(E) = 2/12 = 1/6

Hence, the probability of selecting that the month is March or October , P(E) = 1/6  .

HOPE THIS ANSWER WILL HELP  YOU…

Answer 2

$\underline{\underline{\red{\huge\mathfrak{ANSWER:}}}}$

✔️✔️Option B✔️✔️

✔️✔️$\huge\frac{1}{6}$✔️✔️

$\textbf{EXPLAINATION}$

$\large = >probablity = \frac{no.of \: favourable \: outcomes}{no.of \: total \: outcomes}$

$\large = > \frac{2}{12} \\ \\ = > \frac{1}{6}$

therefore the answer is

✔️✔️option B ✔️✔️

✔️✔️1/6✔️✔️