A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
Answer
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Hint: In this question, we will use the basic probability formula to find the number of black balls in the bag. Probability of A is defined as $P(A)=\dfrac{No.of\text{ }favourable\text{ }outcomes}{Total\text{ }no.of\text{ }possible\text{ }outcomes}$
Complete step-by-step answer:
Given that a bag contains 15 white balls and some black balls.
Let the number of black balls in the bag is x.
According to the question, we know that the probability of drawing a black ball from the bag is thrice that of drawing a white ball from the bag.
That means if we assume the probability of drawing a black ball from the bag as $P(B)$ then probability of drawing a white ball from the bag will be $P(W)$, $P(B)$=$3P(W)\cdot \cdot \cdot \cdot \cdot (1)$
So, the total number of possible outcomes=total number of black balls + the total number of white balls.
We assumed the total number of black balls as x and the total number of white balls as 15. So, the total possible outcomes=15+x.
Number of favourable outcomes of a drawing white ball is 15 whereas favourable outcomes of a drawing black ball=x.
Now, the probability of drawing a white ball is,
$P(W)=\dfrac{No.of\text{ }favourable\text{ }outcomes\text{ }of\text{ }drawing\text{ }white\text{ }balls}{Total\text{ }no.of\text{ }possible\text{ }outcomes}$
$P(W)=\dfrac{15}{15+x}$
Now, the probability of drawing a black ball is,
$P(B)=\dfrac{No.of\text{ }favourable\text{ }outcomes\text{ }of\text{ }drawing\text{ }black\text{ }balls}{Total\text{ }no.of\text{ }possible\text{ }outcomes}$
$P(B)=\dfrac{x}{15+x}$
From 1, $P(B)$=$3P(W)$ . So, $\dfrac{x}{15+x}=3\times \left( \dfrac{15}{15+x} \right)$
Now, we solve for x.
We cancel 15+x term in the denominator in both sides we will get,
$x=3\times 15$
$x=45$
So, the total number of black bags in the bag is 45.
Note: Probability of an event lies between 0 and 1 both 0 and 1 are inclusive. Some black balls in the question implies that there are a countable number of black balls in the bag. It is important to remember the basic formulae of probability.
Complete step-by-step answer:
Given that a bag contains 15 white balls and some black balls.
Let the number of black balls in the bag is x.
According to the question, we know that the probability of drawing a black ball from the bag is thrice that of drawing a white ball from the bag.
That means if we assume the probability of drawing a black ball from the bag as $P(B)$ then probability of drawing a white ball from the bag will be $P(W)$, $P(B)$=$3P(W)\cdot \cdot \cdot \cdot \cdot (1)$
So, the total number of possible outcomes=total number of black balls + the total number of white balls.
We assumed the total number of black balls as x and the total number of white balls as 15. So, the total possible outcomes=15+x.
Number of favourable outcomes of a drawing white ball is 15 whereas favourable outcomes of a drawing black ball=x.
Now, the probability of drawing a white ball is,
$P(W)=\dfrac{No.of\text{ }favourable\text{ }outcomes\text{ }of\text{ }drawing\text{ }white\text{ }balls}{Total\text{ }no.of\text{ }possible\text{ }outcomes}$
$P(W)=\dfrac{15}{15+x}$
Now, the probability of drawing a black ball is,
$P(B)=\dfrac{No.of\text{ }favourable\text{ }outcomes\text{ }of\text{ }drawing\text{ }black\text{ }balls}{Total\text{ }no.of\text{ }possible\text{ }outcomes}$
$P(B)=\dfrac{x}{15+x}$
From 1, $P(B)$=$3P(W)$ . So, $\dfrac{x}{15+x}=3\times \left( \dfrac{15}{15+x} \right)$
Now, we solve for x.
We cancel 15+x term in the denominator in both sides we will get,
$x=3\times 15$
$x=45$
So, the total number of black bags in the bag is 45.
Note: Probability of an event lies between 0 and 1 both 0 and 1 are inclusive. Some black balls in the question implies that there are a countable number of black balls in the bag. It is important to remember the basic formulae of probability.
Last updated date: 23rd Sep 2023
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