Question

A die is thrown once. The probability of getting a number which is not a factor of 36 is (a) 1/6 (b) 1/3 (c) 1/5 (d) 5/6​

Answer 1

Answer:

a) 1/6

Total no. of outcomes is 6 (1, 2,3,4,5,6)

No.s which is not factor of 36 : 1 (5)

ie, Probability : 1/6.

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

A die is thrown once. The probability of getting a number which is not a factor of 36 is

$\displaystyle \sf{ (a) \: \: \frac{1}{6} }$

$\displaystyle \sf{ (b) \: \: \frac{1}{3} }$

$\displaystyle \sf{ (c) \: \: \frac{1}{5} }$

$\displaystyle \sf{ (d) \: \: \frac{5}{6} }$

EVALUATION

Here it is given that a die is thrown once

Then outcomes are 1 , 2 , 3 , 4 , 5 , 6

So total number possible outcomes = 6

Let A be the event that of getting a number which is not a factor of 36

Now the number which is not a factor of 36 among 1 , 2 , 3 , 4 , 5 , 6 is 5

So total number of possible outcomes for the event A is 1

Hence the required probability

= P(A)

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

$\displaystyle \sf{ = \frac{1}{6} }$

FINAL ANSWER

Hence the correct option is

$\displaystyle \sf{ (a) \: \: \frac{1}{6} }$

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