Mix Examples - Probability Questions in English

(A) $(i)$ Total number of balls in the bag $= 3 + 5 = 8$.
The probability of drawing a red ball is given by the formula: $P(\text{red}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
$P(\text{red}) = \frac{3}{8}$.
$(ii)$ The probability of not drawing a red ball is the complement of drawing a red ball.
$P(\text{not red}) = 1 - P(\text{red})$.
$P(\text{not red}) = 1 - \frac{3}{8} = \frac{8 - 3}{8} = \frac{5}{8}$.

(A) Total number of marbles $= 5 + 8 + 4 = 17$.
$(i)$ Number of red marbles $= 5$.
Probability of getting a red marble $= \frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}} = \frac{5}{17}$.
$(ii)$ Number of white marbles $= 8$.
Probability of getting a white marble $= \frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}} = \frac{8}{17}$.
$(iii)$ Number of green marbles $= 4$.
Probability of getting a green marble $= \frac{4}{17}$.
Probability of not getting a green marble $= 1 - P(\text{green}) = 1 - \frac{4}{17} = \frac{13}{17}$.

(C) $(i)$ Total number of bulbs $= 20$.
Total number of defective bulbs $= 4$.
$P(\text{defective bulb}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = \frac{1}{5}$.
$(ii)$ After drawing one non-defective bulb,the remaining total number of bulbs $= 20 - 1 = 19$.
The number of non-defective bulbs remaining $= (20 - 4) - 1 = 15$.
$P(\text{non-defective bulb}) = \frac{\text{Number of non-defective bulbs}}{\text{Remaining total bulbs}} = \frac{15}{19}$.

(N/A) Total number of discs $= 90$.
$(i)$ The two-digit numbers from $1$ to $90$ are $10, 11, \dots, 90$. The total count is $90 - 9 = 81$.
$P(\text{two-digit number}) = \frac{81}{90} = \frac{9}{10}$.
$(ii)$ The perfect square numbers between $1$ and $90$ are $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81$. Total count is $9$.
$P(\text{perfect square}) = \frac{9}{90} = \frac{1}{10}$.
$(iii)$ The numbers between $1$ and $90$ divisible by $5$ are $5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90$. Total count is $18$.
$P(\text{divisible by } 5) = \frac{18}{90} = \frac{1}{5}$.

(A) $(i)$ $Incorrect$
When two coins are tossed,the possible outcomes are $(H, H), (H, T), (T, H),$ and $(T, T)$.
It can be observed that there can be one of each in two possible ways: $(H, T)$ and $(T, H)$.
Therefore,the probability of getting two heads is $\frac{1}{4}$,the probability of getting two tails is $\frac{1}{4},$ and the probability of getting one of each is $\frac{2}{4} = \frac{1}{2}$.
Thus,the probability for each of these outcomes is not $\frac{1}{3}$.
$(ii)$ $Correct$
When a die is thrown,the possible outcomes are $1, 2, 3, 4, 5,$ and $6$. Out of these,$1, 3, 5$ are $odd$ and $2, 4, 6$ are $even$ numbers.
Since there are $3$ odd numbers out of $6$ total outcomes,the probability of getting an odd number is $\frac{3}{6} = \frac{1}{2}$.

(3) Total number of balls $= 12$.
Total number of black balls $= x$.
The probability of drawing a black ball is $P_1 = \frac{x}{12}$.
If $6$ more black balls are added,the new total number of balls $= 12 + 6 = 18$.
The new number of black balls $= x + 6$.
The new probability of drawing a black ball is $P_2 = \frac{x+6}{18}$.
According to the problem,$P_2 = 2 \times P_1$.
Therefore,$\frac{x+6}{18} = 2 \times \frac{x}{12}$.
$\frac{x+6}{18} = \frac{x}{6}$.
Multiplying both sides by $18$,we get $x + 6 = 3x$.
$2x = 6$,which gives $x = 3$.

(A) Total number of cards in a deck = $52$.
After removing the king,queen,and jack of clubs,the number of remaining cards is $n(S) = 52 - 3 = 49$.
$(i)$ Let $E_1$ be the event of getting a heart.
Since there are $13$ hearts in a deck and none were removed,$n(E_1) = 13$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{13}{49}$.
$(ii)$ Let $E_2$ be the event of getting a king.
There are $4$ kings in a deck. Since the king of clubs was removed,the number of remaining kings is $n(E_2) = 4 - 1 = 3$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{3}{49}$.

(A) Total number of outcomes with numbers $2$ to $101$ is $n(S) = 101 - 2 + 1 = 100$.
$(i)$ Let $E_1$ be the event of selecting a card which is an even number. The even numbers are $\{2, 4, 6, \dots, 100\}$.
This is an arithmetic progression where $a = 2$,$d = 2$,and $l = 100$.
Using $l = a + (n - 1)d$,we get $100 = 2 + (n - 1)2$,which implies $98 = 2(n - 1)$,so $n - 1 = 49$,and $n = 50$.
Thus,$n(E_1) = 50$.
The probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{50}{100} = \frac{1}{2}$.
$(ii)$ Let $E_2$ be the event of selecting a card which is a square number.
The square numbers between $2$ and $101$ are $\{4, 9, 16, 25, 36, 49, 64, 81, 100\}$.
These correspond to $\{2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2, 10^2\}$.
Thus,$n(E_2) = 9$.
The probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{9}{100}$.

(D) Total number of slips in the third box,$n(S) = 25 + 50 = 75$.
From the given information:
Number of slips marked $Rs. 1$ in Box $A = 19$.
Number of slips marked $Rs. 5$ in Box $A = 25 - 19 = 6$.
Number of slips marked $Rs. 1$ in Box $B = 45$.
Number of slips marked $Rs. 13$ in Box $B = 50 - 45 = 5$.
Total number of slips marked other than $Rs. 1 = 6 + 5 = 11$.
Therefore,the required probability $= \frac{\text{Number of slips marked other than } Rs. 1}{\text{Total number of slips}} = \frac{11}{75}$.

(N/A) Total number of bulbs, $n(S) = 24$.
Number of defective bulbs $= 6$.
Number of non-defective (good) bulbs $= 24 - 6 = 18$.
Part $1$: Probability that the bulb is not defective:
$P(\text{not defective}) = \frac{\text{Number of good bulbs}}{\text{Total number of bulbs}} = \frac{18}{24} = \frac{3}{4}$.
Part $2$: If the first bulb selected is defective and not replaced:
Remaining total number of bulbs $= 24 - 1 = 23$.
Remaining number of defective bulbs $= 6 - 1 = 5$.
Probability that the second bulb is defective $= \frac{\text{Remaining defective bulbs}}{\text{Remaining total bulbs}} = \frac{5}{23}$.

(N/A) Given,total number of mobile phones $n(S) = 48$.
$(i)$ Let $E_{1}$ be the event that Varnika will buy a mobile phone.
Varnika buys only if it is a good mobile.
Therefore,$n(E_{1}) = 42$.
$P(E_{1}) = \frac{n(E_{1})}{n(S)} = \frac{42}{48} = \frac{7}{8}$.
$(ii)$ Let $E_{2}$ be the event that the trader will buy a mobile phone.
The trader will buy only if it has no major defects.
This means the trader will buy the $42$ good phones and the $3$ phones with minor defects.
Therefore,$n(E_{2}) = 42 + 3 = 45$.
$P(E_{2}) = \frac{n(E_{2})}{n(S)} = \frac{45}{48} = \frac{15}{16}$.

(D) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $E$ be the event that the sum of the numbers on the two dice is a multiple of $3$ or $4$. The possible sums range from $2$ to $12$. Multiples of $3$ are $3, 6, 9, 12$. Multiples of $4$ are $4, 8, 12$.
Thus,we need the sum to be in the set $\{3, 4, 6, 8, 9, 12\}$.
Outcomes with sum $3$: $(1,2), (2,1)$
Outcomes with sum $4$: $(1,3), (2,2), (3,1)$
Outcomes with sum $6$: $(1,5), (2,4), (3,3), (4,2), (5,1)$
Outcomes with sum $8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$
Outcomes with sum $9$: $(3,6), (4,5), (5,4), (6,3)$
Outcomes with sum $12$: $(6,6)$
Total favorable outcomes = $2 + 3 + 5 + 5 + 4 + 1 = 20$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{36} = \frac{5}{9}$.

(C) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The possible outcomes are:
$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$
$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$
$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$
$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$
Let $H$ be the event that the sum of the numbers on the two dice is a factor of $18$. The factors of $18$ are $1, 2, 3, 6, 9, 18$. Since the minimum sum is $2$ and the maximum sum is $12$,the possible sums that are factors of $18$ are $2, 3, 6, 9$.
Favorable outcomes for each sum:
Sum $= 2: (1,1)$
Sum $= 3: (1,2), (2,1)$
Sum $= 6: (1,5), (2,4), (3,3), (4,2), (5,1)$
Sum $= 9: (3,6), (4,5), (5,4), (6,3)$
Total number of favorable outcomes $= 1 + 2 + 5 + 4 = 12$.
Probability $P(H) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{36} = \frac{1}{3}$.

(D) Area of the rectangle $=$ Length $\times$ Breadth
$= 28 \, cm \times 22 \, cm = 616 \, cm^2$
For the circle,radius $r = 7 \, cm$
Area of the circle $= \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \, cm^2$
The probability that a ball hitting the rectangle hits the circle is given by the ratio of the area of the circle to the area of the rectangle.
Required Probability $= \frac{\text{Area of the circle}}{\text{Area of the rectangle}}$
$= \frac{154}{616} = \frac{1}{4}$
Thus,the probability is $\frac{1}{4}$.

(B) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability that the sum $S$ of the numbers on the two dice lies between $5$ and $9$,which means $5 < S < 9$.
This implies the sum $S$ can be $6, 7,$ or $8$.
- For $S = 6$,the possible outcomes are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,which are $5$ outcomes.
- For $S = 7$,the possible outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$,which are $6$ outcomes.
- For $S = 8$,the possible outcomes are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$,which are $5$ outcomes.
Total favorable outcomes = $5 + 6 + 5 = 16$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{16}{36} = \frac{4}{9}$.