5 green,10 yellow,3 blue and 2 white marbles are put in a bowl. Out of two draws without any replacement find the probability of :
A green and a blue marble in any order.
plz use nCr combination method.
Answers
Answer:
A is answer
Step-by-step explanation:
I HOPE it help full for you
Answer:There are 2 ways of doing this problem. They are as follows:
Method 1:
Total no. Of marbles in the box = 5+4+3=12.
No. Of ways 3 marbles can be drawn from 12 marbles = 12C3 = 12*11*10/6 = 220
Now, drawing 3 marbles of different colors means drawing 1 marble of each of the 3 colors.
No. Of ways a green marble can can draw from 5 green marbles = 5C1 = 5
No. Of ways a yellow marble can be drawn from 4 yellow marbles = 4C1 = 4
No. Of ways a white marble can draw from 3 white marbles = 3C1 = 3
Therefore, the total probability of drawing 3 marbles from 12 marbles = Probability of drawing 1 green, 1 yellow and 1 white marble from the box = 5*4*3/220=3/11 (Since, this is an intersection, hence, the no. Of ways of drawing 1 marble of each color, is multiplied)
Method 2 :
Total no. Of marbles in the box = 5+4+3 =12
Let a green marble be drawn at first, followed by a yellow and a white marble, “without replacement”.
No. Of ways of selecting a green marble from 12 marbles = 5/12
No. Of ways of selecting a yellow marble (12–1) marbles = 4/11 (Since, one marble has already been drawn)
No. Of ways of selecting a white marble (12–2) marbles = 3/10 (Since, 2 marbles have already been drawn)
Now, twist in the solution :
It is not mentioned in the question that which color of marble is drawn at first, so we cannot assume the order to be green -> yellow -> white. Since, there are 3 different colors of marbles, so we can arrange the ways of selecting them 3! Ways. So, the answer should be :
Required Probability = (5/12) * (4/11) * (3/10) *6 =3/11
So see, we get the same answer in both the methods. Hence, clearly, our approach is correct in both the cases.
Step-by-step explanation: