Question

Rohit and Miranda each roll one fair die numbered from 1to6 if P is the probability that Miranda gets a higher number than Rohit then find the value of 120p​

Answer 1

Answer:

The value of 120p is 50

Step-by-step explanation:

since each roll one fair die,

the sample space

S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Let E be the event of getting Miranda gets a higher number than Rohit

E={ (1,2), (1,3), (1,4), (1,5), (1,6)

(2,3), (2,4), (2,5), (2,6)

(3,4), (3,5), (3,6)

(4,5), (4,6)

(5,6) }

n(E)=15

Now,

$P(E)=\frac{n(E)}{n(S)}$

$\implies\:=p=P(E)=\frac{15}{36}$

$\implies\:p=P(E)=\frac{5}{12}$

$120p=120*\frac{5}{12}$

$\implies\:120p=10*5=50$