Question

two dice are thrown together. find the the probability that the no is
1 have sum less than 7
2 have product less than 16 3 is a doublet of odd no
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Answer 1

SEE THE ATTACHMENT BELOW

THANK YOU ❤

Answer 2

$\huge{\bold{\pink{SOLUTION:-}}}$

The outcomes when two dice are thrown together are-

(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)

${\boxed{\mathcal{\red{Total \:no. \:of\: outcomes=36}}}}$

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a) Let A be the event of getting the numbers whose sum is less than 7

The outcomes in favor of event a are-

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

${\boxed{\mathcal{\green{No.\:of \: favourable\: outcomes=15}}}}$

Therefore, P(A)=$\large\ \frac{no \: of \: favourble \: outcomes}{total \: no \: of \: outcomes}$

$= \large\frac{15}{36}$

$= \large \frac{5}{2}$

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b) let B be the event of getting the numbers whose product is less than 163.

The outcomes in favour of event B are-

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

${\boxed{\mathcal{\blue{No.\:of \: favourable\: outcomes=25}}}}$

Therefore,P(A)=$\large\ \frac{no \: of \: favourble \: outcomes}{total \: no \: of \: outcomes}$

$= \frac{25}{36}$

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$\huge{ \mathfrak{ \orange{Hope\:it\:helps...}}}$