Question

A box contains 5 green pencils and 7 yellow pencils. Two pencils are chosen at random from the box without replacement. What is the probability they are both yellow?

Answer 1

To find the probability of two events happening, we must multiply the probability of each separate event happening by the other.

So, for the first time that you pick a pencil, we have to find the probability that it is a yellow pencil. 

Probability = favorable outcomes / total possible outcomes = 7 yellow / 7 yellow + 5 green = 7 yellow pencils / 12 total pencils

Thus, the probability of the first event is 7/12.

According to the question, we want to pick a yellow pencil both times, but we don't replace it.  So, assuming we pick a yellow pencil, we have to subtract that from the next calculation.

Thus, probability of the second event = probable outcomes/ total outcomes = 6 yellow pencils / 6 yellow + 5 green pencils = 6 / 11

Therefore, the second event's probability is 6/11.

To find the total probability, we must multiply the two fractions together.

7 / 12 * 6 / 11 = 42 / 132

But, we want to simplify this fraction, so:

42 / 132 = 42 ÷ 6 / 132 ÷ 6 = 7/22

Thus, the probability is 7/22, or approximately 32%.

Hope this helps! :)

Answer 2

Hello!

A box contains 5 green pencils and 7 yellow pencils. Two pencils are chosen at random from the box without replacement. What is the probability they are both yellow?

Data:

total pencils: 12

green pencils: 5

yellow pencils: 7  

We can choose two pieces between the twelve, therefore, we have a simple combination:

$C_{12,2} = \dfrac{12!}{2!(12-2)!} = \dfrac{12!}{2!10!} =\dfrac{12*11*\diagup\!\!\!\!\!\!10!}{2!\diagup\!\!\!\!\!\!10!}=\dfrac{132}{2}=\boxed{66\:different\:ways}$

Now, we choose two among the seven yellow:

$C_{7,2} = \dfrac{7!}{2!(7-2)!} = \dfrac{7!}{2!5!} =\dfrac{7*6*\diagup\!\!\!\!\!5!}{2!\diagup\!\!\!\!\!5!}=\dfrac{42}{2}= \boxed{21\:different\:ways}$

Therefore, the probability of both being yellow:

$\dfrac{21}{66}\frac{\div3}{\div3}\to\boxed{\boxed{\frac{7}{22}\:different\:ways}}\:\:or\:\:\boxed{\boxed{\approx 32\:\%\:different\:ways}} \end{array}}\qquad\quad\checkmark$

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I Hope this helps, greetings ... Dexteright02! =)