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Probability — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsProbability3 Marks
Question
Two different dice are thrown together. Find the probability that the numbers obtained
$(i)$ have a sum less than $7$
$(ii)$ have a product less than $16$
$(iii)$ is a doublet of odd numbers.
✓

Answer

We know that the total number of outcome when two dice are thrown to gether is $36$
$(i)$ Have a sum less than $7$
The favourable outcomes are
$=\{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),$
$(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),$
$(4,1),(4,2),(5,1)\}$
Number of favourable outcomes $=15$
So, probability of getting a sum less than $7$
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{15}{36}$
$=\frac{5}{12}$
$(ii)$ Have a product less than $16 $
The favourable outcomes are
$=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),$
$(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),$
$(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(5,1),$
$(5,2),(5,3),(6,1),(6,2)\}$
Number of favourable outcomes $=25$
So, probability will be
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{25}{36}$
$(iii) A$ doublet of odd numbers.
The favourable outcomes are
$=\{(1,1),(3,3),(5,5)\}$
Number of favourable outcomes $=3$
So, probability will be
$=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{36}$
$=\frac{1}{12}$

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