(i) the same day?
(ii) consecutive days?
(iii) different days?
Since there are 5 days and both can go to the shop in 5 ways each so,
The total number of possible outcomes = 5×5 = 25
(i) the same day?
The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)
So, P (both visiting on the same day) = $\dfrac{5}{25} = \dfrac{1}{5}$
(ii) consecutive days?
The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)
So, P(both visiting on the consecutive days) = $\dfrac{8}{25}$
(iii) different days?
P (both visiting on different days) = 1-P (both visiting on the same day)
So, P (both visiting on different days) = 1- $\dfrac{1}{5} = \dfrac{4}{5}$
(i) even?
(ii) 6?
(iii) at least 6?
The table will be as follows:
So, the total number of outcomes = 6×6 = 36
(i) even?
E (Even) = 18
P (Even) = $\dfrac{18}{36} = \dfrac{1}{2}$
(ii) 6?
E (sum is 6) = 4
P (sum is 6) = $\dfrac{4}{36} = \dfrac{1}{9}$
(iii) at least 6?
E (sum is atleast 6) = 15
P (sum is atleast 6) = $\dfrac{15}{36} = \dfrac{5}{12}$
It is given that the total number of red balls = 5
Let the total number of blue balls = x
So, the total no. of balls = x+5
P(E) = $\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$ |
∴ P (drawing a blue ball) = $\dfrac{x}{x+5}$ ——–(i)
Similarly,
P (drawing a red ball) = $\dfrac{5}{x+5}$ ——–(ii)
From equations (i) and (ii)
P (drawing a blue ball) = 2 P (drawing a red ball)
$\dfrac{x}{x+5}$ = 2 × $\dfrac{5}{x+5}$
x = 10
So, the total number of blue balls = 10
Total number of black balls = x
Total number of balls = 12
P(E) = $\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$ |
P (getting black balls) = $\dfrac{x}{12}$ ——————-(i)
Now, when 6 more black balls are added,
Total balls become = 18
∴ Total number of black balls = x+6
Now, P (getting black balls) = $\dfrac{(x+6)}{18}$ ——————-(ii)
It’s given that the probability of drawing a black ball now is double of what it was before.
eqn (ii) = 2 × eqn (i)
$\dfrac{(x+6)}{18}$ = 2 × $\dfrac{x}{12}$
x + 6 = 3x
2x = 6
∴ x = 3
Total marbles = 24
Let the total green marbles = x
So, the total blue marbles = 24-x
P(getting green marble) = $\dfrac{x}{24}$
From the question, $\dfrac{x}{24} = \dfrac{2}{3}$
So, the total green marbles = 16
And, the total blue marbles = 24-16 = 8