A box contains 5 red and k blue balls. A ball is selected at random from the box. If the probability of selecting a blue ball is 23, find the value of k ?
Answers
Answer:
10
Step-by-step explanation:
Let the number of blue balls =x
Total number of balls =5+x
Probability for a red ball = x+5
Probability for a blue ball = 5+x
According to the question
5+x
=2(x+5)⇒x=10
∴ Number of blue balls =10.
Step-by-step explanation:
Given :-
A box contains 5 red and k blue balls. A ball is selected at random from the box. The probability of selecting a blue ball is 23.
Correction :-
The probability of selecting a blue ball is 2/3
To find :-
Find the value of k ?
Solution :-
Given that
Number of red balls in a box = 5
Number of favourable outcomes to red ball = 5
Number of blue balls in the box = k
Number of favourable outcomes to blue ball = k
Total number of all balls in the box = (5+k)
Total number of all possible outcomes = (5+k)
We know that
The probability of getting an even P(E) =
Number of favourable outcomes / Total number of all possible outcomes
Now,
Probability of getting a blue ball = P(B)
=> Number of favourable outcomes to blue ball / Total number of all possible outcomes
=> P(B) = k / (5+k)
According to the given problem
Probability of getting a blue ball = 2/3
=> k / (5+k) = 2/3
On applying cross multiplication then
=> 3k = 2(5+k)
=> 3k = 10+2k
=> 3k-2k = 10
=> k = 10
Therefore, k = 10
Number of blue balls = 10
Answer :-
The value of k for the given problem is 10
Note :-
If the Probability of getting a blue ball = 23 then we get the value of k as negative ,So it is not possible for the number of blue balls ,it must be a positive number.
Used formulae:-
→ The probability of getting an even P(E) =
Number of favourable outcomes / Total number of all possible outcomes