Question

an unbiased die is tossed twice .find the probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss

Answer 1

When a dice is tossed , total possible outcome = {1,2,3,4,5,6}

When , Dice is tossed twice Total Possible Outcome = 6²=36

Favorable Outcome = A={4,5,6} and B={1,2,3,4}

Probability of an event = $\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}$

P(A)= $\frac{3}{6}=\frac{1}{2}$

P(B)=  $\frac{4}{6}=\frac{2}{3}$

Probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss

     = P(A) × P(B)

        = $\frac{1}{2} \times\frac{2}{3}\\\\= \frac{1}{3}$                    

Answer 2

Answer:

When a dice is tossed , total possible outcome = {1,2,3,4,5,6}

When , Dice is tossed twice Total Possible Outcome = 6²=36

Favorable Outcome = A={4,5,6} and B={1,2,3,4}

Probability of an event = \frac{\text{Total favorable outcome}}{\text{Total possible outcome}}

Total possible outcome

Total favorable outcome

P(A)= \frac{3}{6}=\frac{1}{2}

6

3

=

2

1

P(B)= \frac{4}{6}=\frac{2}{3}

6

4

=

3

2

Probability of getting a 4,5,6 on the first toss and a 1,2,3,4 on the second toss

= P(A) × P(B)

= \begin{lgathered}\frac{1}{2} \times\frac{2}{3}\\\\= \frac{1}{3}\end{lgathered}

2

1

×

3

2

=

3

1

hope it helps you