Chapter 14: Probability

Ans: If Shyam and Ekta are visiting a particular shop on the same week which starts from Tuesday and ends on Saturday then the total days in the week $=5$

Now, if each of them visiting the shop is equally likely then,

(The outcome of the event of Shyam visiting the shop) $=5$

(The outcome of the event of Ekta visiting the shop) $=5$

Thus, the total outcome of events

$\begin{aligned} \Rightarrow \mathrm{P}(\text { Total outcome }) &=5 \times 5 \\ &=25 \end{aligned}$

Ans: Probability of both of them reaching the shop on the same day:

$\mathrm{P}($ Same day $)=(\mathrm{T}, \mathrm{T}),(\mathrm{W}, \mathrm{W}),(\mathrm{Th}, \mathrm{Th}),(\mathrm{F}, \mathrm{F}),(\mathrm{S}, \mathrm{S})$.

$=5$

$P$ (reaching on the same day) $=\frac{P(\text { sameday })}{P(\text { Total outcomes })}$

$ \begin{aligned} &=\frac{5}{25} \\ &=\frac{1}{5} \end{aligned} $

Answer $=\frac{1}{5}$ 

Ans: The probability of both of them reaching on consecutive days:

$\mathrm{P}($ Consecutive days $)=(\mathrm{T}, \mathrm{W}),(\mathrm{W}, \mathrm{Th}),(\mathrm{Th}, \mathrm{F}),(\mathrm{F}, \mathrm{S}),(\mathrm{W}, \mathrm{T}),(\mathrm{Th}, \mathrm{W}),(\mathrm{F}, \mathrm{Th})$, $(\mathrm{S}, \mathrm{F})$

$=8$

$\mathrm{P}$ (reaching on consecutive days) $=\frac{\mathrm{P}(\text { consecutive days })}{\mathrm{P}(\text { Total outcomes })}$

$ =\frac{8}{25} $

Answer $=\frac{8}{25}$

Ans: Let $P$ (both of them reaching on same days) $+P$ (both of them not reaching on the same days) $=1$

$\frac{1}{5}+\mathrm{P}$ (both of them reaching not on same days) $=1$

$\mathrm{P}$ (both of them reaching not on same days) $=1-\frac{1}{5}$

$ =\frac{4}{5} $

Answer $=\frac{4}{5}$

Ans:

i) $\mathrm{P}$ (getting the total score as even) $=\frac{\text { ways of getting even sum }}{\text { total ways }}$ 

$ =\frac{18}{36} $

ii) $P$ ( getting total as 6 ) $=\frac{\text { ways of getting total as } 6}{\text { total ways }}$

$ =\frac{5}{36} $

iii) $\mathrm{P}$ ( getting total sum as 6 or less than 6 ) $=$ ways of getting sum equal orless than 6

$ \begin{aligned} \text { total ways } &=\frac{15}{36} \end{aligned} $

Ans: A bag of 5 red balls and x blue balls (assume) will be having total $5+x$ balls.

$P($ getting a blue ball $)=\frac{x}{5+x}$

$P($ getting a red ball $)=\frac{5}{5+x}$

If the probability of getting a blue ball is double the probability of getting a red ball then

$2 \mathrm{P}$ (getting a blue ball) $=\mathrm{P}$ (getting a red ball)

$ \begin{aligned} 2\left(\frac{x}{5+x}\right) &=\frac{5}{5+x} \\ 2 x^{2}+5 x-25 &=0 \end{aligned} $

On solving the following equation

$ \mathrm{x}=10,-5 $

Number of blue balls cannot be negative thus $x=10$.

$ \text { Answer }=10 $

Ans:

i) If a box of 12 balls has $x$ black balls and $12-\mathbf{x}$ 0ther balls then

$\mathrm{P}$ (getting a black ball) $=\frac{\text { numberof black balls }}{\text { total balls }}$

$ =\frac{\mathrm{x}}{12} $

ii) If 6 more balls are added then the total number of black balls $=x+6$

And the total number of balls $=12+6=18$

So, $\mathrm{P}$ (getting a black ball) $=\frac{\text { number of black balls }}{\text { total balls }}$

$ =\frac{x+6}{18} $

Now, if the probability of getting a black ball is double of before then, $2 \mathrm{P}$ ( getting a black ball before) $=\mathrm{P}$ (getting a black ball)

Therefore, $2\left(\frac{x}{12}\right)=\frac{x+6}{18}$

$ 36 x-12 x=72 $

$ \begin{aligned} & 24 x=72 \\ & \mathrm{x}=3 \end{aligned} $

Thus, the number of black balls before was 3 .

Answer $=3$

Answer: The total number of marbles $=24$

Let the total number of green marbles be $x$ then

The total number of blue marbles $=24-\mathrm{X}$

$\mathrm{P}$ (getting a green marble) $=\frac{\text { numberof green marbles }}{\text { total number of marbles }}$

$ =\frac{x}{24} $

Given, $P$ (getting a green marble) $=\frac{2}{3}$

Putting the value, $\frac{x}{24}=\frac{2}{3}$

$ \begin{aligned} 3 x &=48 \\ x &=16 \end{aligned} $

Therefore, if the number of green marbles is 16 then, the number of blue marbles is, $24-16=8$

Answer $=8$