Question

A jar contain 24 marbles some are green and other are blue if a marble is drawn at random from the jar the probability that it is green is 1 / 3 find the number of balls of each type​

Answer 1

Given that:

• A jar contain 24 marbles some are green and other are blue.
• A marble is drawn at random from the jar the probability that it is green is 1/3.

To Find:

• The number of balls of each type.

We know that:

• P(E) = F/T

Where,

• P(E) = Probability of an event
• F = Favourable outcomes
• T = Total outcomes

Let us assume:

• The number of green ball be x.
• Number of blue ball = 24 - x

Finding the number of balls of each type:

According to the question.

⟶ x/24 = 1/3

Cross multiplication.

⟶ 3x = 24

⟶ x = 24/3

⟶ x = 8

Number of green ball = x = 8

Number of blue ball = 24 - x = 24 - 8 = 16

Hence,

• There are 8 green balls and 16 blue balls.

Answer 2

$\underline{\textsf{\textbf{Correct\:Question\::}}}$

• A jar contain 24 marbles some are green and other are blue if a marble is drawn at random from the jar the probability that it is green is 1/3 find the number marbles of each type.

$\underline{\textsf{\textbf{Required\:Answer\::}}}$

• There are 8 green marbles and 16 blue marbles in a jar.

$\underline{\textsf{\textbf{Step\:by\:step\:explanation\:}}}$

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• A jar contains 24 marbles, some are green and some are blue.
• Probability of green marble is 1/3 if a marble is drawn from jar at random.
• We had to find out the number marbles of each type.
• Let number of green marbles be y.
• So, number of blue marbles = 24 - y.

$\underline{\textsf{\textbf{Using\:formula\::}}}$

$\qquad\red\bigstar\:{\tiny{\underline{\boxed{\bf{\green{Probability \:of\:an\:event[P(E)] = \dfrac{Favorable\:outcomes}{Total\:outcomes}}}}}}}$

$\underline{\tt{\bigstar\:According\:to\:the\:question\::-}}$

$\dashrightarrow\qquad\sf P(E)_{(green\:marble)} = \dfrac{1}{3}$

$\dashrightarrow\qquad\sf \dfrac{y}{24} = \dfrac{1}{3}$

$\qquad \qquad\tiny \dag \: {\underline{\frak{By\:cross\;multiplication\::}}}$

$\dashrightarrow\qquad\sf 3y = 24$

$\dashrightarrow\qquad\sf y = \dfrac{24}{3}$

$\dashrightarrow\qquad\sf y = \dfrac{\cancel{24}}{\cancel{3}}$

$\dashrightarrow\qquad{\boxed{\frak{\pink{y = 8}}}}\:\purple\bigstar$

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$\underline{\textsf{\textbf{Therefore\::}}}$

• Number of green marbles = y = 8
• Number of blue marbles = 24 - y = 24 - 8 = 16

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$\small\therefore\:{\underline{\sf{Hence,\:there \:are\: \bf{8}\:\sf{green\:marbles\: and}\: \bf{16}\: \sf{blue \:marbles\: in\: a\: jar\:respectively}.}}}$