Question

in a single throw of two dice find the probability that neither a doublet nor a total of 9 will appear

Answer 1

Answer:

$\frac{13}{18}$

Step-by-step explanation:

Two dices are rolled .

So, outcomes :

{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}

So, total outcomes = 36

So, to find probability of getting a doublet or a total of 9

Favorable events = {1,1} ; {2,2} ; {3,3} ; {3,6} ; {4,4} ; {4,5} ;  {5,4} ; {5,5} ;{6,3} ;  {6,6} = 10

So, probability of of getting a doublet or a total of 9 = $\frac{10}{36}$

So,  the probability that neither a doublet nor a total of 9 will appear :

$1 - \frac{10}{36}$

$\frac{13}{18}$

hence the probability hat neither a doublet nor a total of 9 will appear is $\frac{13}{18}$

Answer 2

Answer:

$\frac{7}{9}$