Solveeit Logo

Question

Question: A pot has 2 white, 6 black, 4 grey, 8 green balls. If one ball is picked randomly from the pot, what...

A pot has 2 white, 6 black, 4 grey, 8 green balls. If one ball is picked randomly from the pot, what is the probability of it being black or green?
A. 34\dfrac{3}{4}
B. 110\dfrac{1}{{10}}
C. 43\dfrac{4}{3}
D. 710\dfrac{7}{{10}}

Explanation

Solution

Here, we can use the formula P(A+B)P\left( {A + B} \right) of probability, so we find the required answer. Formula we will use here is P(A+B)=P(A)+P(B)P\left( {A + B} \right) = P\left( A \right) + P\left( B \right) & then we will put the values as determined from the question.

Complete step-by-step answer:
Given data in the question,
Total no. of balls in the pot is = 20
Let’s consider that A and B denotes the event that the ball drawn is black or the ball drawn is green. Then P(A) & P(B) denote probability of event A & B respectively.
So, applying the formula of addition of probabilities of two independent events is P(A+B)=P(A)+P(B)P\left( {A + B} \right) = P\left( A \right) + P\left( B \right)
As there are 6 black balls out of 20 , so P(A)=620=310 P\left( A \right) = \dfrac{6}{{20}} = \dfrac{3}{{10}}{\text{ }}
& as there are 8 green balls out of 20, so  P(B)=820=410{\text{ }}P\left( B \right) = \dfrac{8}{{20}} = \dfrac{4}{{10}}
Putting the values P(A+B)=P(A)+P(B)P\left( {A + B} \right) = P\left( A \right) + P\left( B \right) [As the event is mutually exclusive]
=310+410=3+410=710= \dfrac{3}{{10}} + \dfrac{4}{{10}} = \dfrac{{3 + 4}}{{10}} = \dfrac{7}{{10}}

So, the correct answer is “Option D”.

Additional information: Key concept in this numerical is of Mutually Exclusive event:
Two events are said to be mutually exclusive events when both cannot occur at the same time. In probability, the outcomes of an experiment are what we call the events. Some of these events have relations with other events. In other words, we say that some events affect the occurrence of other events. Here we will study one type of events that we call mutually exclusive events.

Note: We can use this type of question using P(A+B)P\left( {A + B} \right) and here the events are mutually exclusive so P(A+B)=P(A)+P(B)P\left( {A + B} \right) = P\left( A \right) + P\left( B \right). Calculation should be done attentively to avoid silly mistake instead of having concept of Mutually exclusive events & its formula