Question

The faces of a die bear the numbers 1, 3 , 5, 7, 9, 11. The die is rolled. Find the probability of getting a

perfect square number on the upper face of the dice.

Here sample space S = {1, 3, 5, 7, 9, 11}

n(S) = 6

let A be event of getting a perfect square number then A = { _____________________ }

n(A) = ________

P(A) =



= _________ = ________​

Answer 1

Answer:

A= 3,9

n(A) = 2

P(A)= n(A)/ n(s)

P(A)=2/6

=1/3

Therefore, probability of getting a perfect square is 1/3

HOPE IT HELPS :)

Answer 2

Given:

We have faces of die with the numbers1,3,5,9,11.

To Find:

We have to find the probabilty of getting perfect square?

Step-by-step explanation:

Probability:

• Probability of an event is known as possibllity of occuring of that event.

• we can find the probability of any event by using the formula that is no. of favorable outcomes divided by the total number of outcomes.

       Now we have that faces of a die shows the numbers 1,3,5,7,9,11

       Thus sample space is witten as

• Sample space={1,3,5,7,9,11}

• Total number of terms is 6.

• Now Let A by the event of getting perfect square number

• We have perfect square numbers in given sample space are 1,9

• Hence A={1,9}

        n(A)= no. of elements in A= 2

• Hence probability of getting a perfect square number is given by the formula

       $P(E)=\frac{\textrm{total no. of outcome}}{\textrm{total no.}} \\\\P(A)=\frac{2}{6} =\frac{1}{3}$

Hence, probabilty of A is P(A)=$\frac{1}{3}$