Exercise 11.1 - Chapter 11 - Probability Distributions - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Probability Distributions
Ex 11.1
Question 1.
Suppose $\mathrm{X}$ is the number of tails occurred when three fair coins are tossed once simultaneously. Find the values of the random variable $\mathrm{X}$ and number of points in its inverse images.
Solution:
Let $\mathrm{X}$ is the random variable denotes the number of tails when three coins are tossed simultaneously.
Sample space $\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{HTT}, \mathrm{THH}$, THT, TTH, TTT $\}$
$\therefore$ ' $\mathrm{X}$ ' takes the values $0,1,2,3$
i.e., $\mathrm{X}(\mathrm{HHH})=0 ; \mathrm{X}(\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH})=1 ; \mathrm{X}(\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH})=2 ; \mathrm{X}(\mathrm{TTT})=3$

Question 2.
In a pack of 52 playing cards, two cards are drawn at random simultaneously. If the number of black cards drawn is a random variable, find the values of the random variable and number of points in its inverse images.
Solution:
Total number of playing cards $=52$
Number of Black cards $=26$
Number of Non-black (or) Red cards $=26$
Let ' $\mathrm{X}$ ' be the random variable denotes the number of black cards. Since two black cards are drawn, ' $\mathrm{X}$ ' takes the values $0,1,2$
$\mathrm{X}($ Non-black Cards $)=\mathrm{X}\left(26 \mathrm{C}_1 \times 25 \mathrm{C}_1\right)=\mathrm{X}(650)=0$
$\mathrm{X}(1$ Black Card $)=X\left(26 \mathrm{C}_1 \times 26 \mathrm{C}_0\right)=\mathrm{X}(26)=1$
$\mathrm{X}(2$ Black Cards $)=X\left(26 \mathrm{C}_1 \times 25 \mathrm{C}_1\right)=\mathrm{X}(650)=2$

Question 3.
An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.
Solution:
Number of mangoes $=5$

Number of Apples $=4$
Total number of fruits $=9$
Let ' $X$ ' be the random variable denotes the number of apples taken, then it takes the values $0,1,2,3$
$\mathrm{X}(\mathrm{MMM})=0$
$\mathrm{X}(\mathrm{AMM}$ (or) MAM (or) MMA) $=1$
$\mathrm{X}(\mathrm{AAM}$ (or) $\mathrm{AMA}$ (or) $\mathrm{MAA})=2$
$\mathrm{X}(\mathrm{AAA})=3$

Question 4.
Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win ₹ 15 for each red ball selected and we lose ₹ 10 for each black ball selected. $\mathrm{X}$ denotes the winning amount, then find the values of $\mathrm{X}$ and number of points in its inverse images.
Solution:
Number of red balls $=6$
Number of black balls $=8$
' $\mathrm{X}$ ' is the random variable denotes the winning amount.
$\therefore$ The values of ' $\mathrm{X}$ ' are $0,15,30$
i.e., $X(\mathrm{BB})=0$
$\mathrm{X}(\mathrm{RB})=15+0=15$
$X(R R)=15+15=30$

Question 5.
A six sided die is marked ' 2 ' on one face, ' 3 ' on two of its faces, and ' 4 ' on remaining three faces. The die is thrown twice. If $\mathrm{X}$ denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Solution:
Six sided die marked ' 2 ' on one face, ' 3 ' on two faces and ' 4 ' on three faces.
When it is thrown twice, we get 36 sample points.
' $X$ ' denotes sum of the face numbers and the possible values of ' $X$ ' are $4,5,6,7$ and 8
For $X=4$, the sample point is $(2,2)$
For $\mathrm{X}=5$, the sample points are $(2,3),(3,2)$
For $\mathrm{X}=6$, the sample points are $(3,3),(2,4),(4,2)$
For $\mathrm{X}=7$, the sample points are $(3,4),(4,3)$
For $\mathrm{X}=8$, the sample point is $(4,4)$