Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 12)
12.
A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Answer: Option
Explanation:
Let S be the sample space.
| Then, n(S) | = number of ways of drawing 3 balls out of 15 | |||
| = 15C3 | ||||
|
||||
| = 455. |
Let E = event of getting all the 3 red balls.
 n(E) = 5C3 = 5C2 = |
(5 x 4) | = 10. |
| (2 x 1) |
| P(E) = |
n(E) | = | 10 | = | 2 | . |
| n(S) | 455 | 91 |
Discussion:
33 comments Page 3 of 4.
Shiva said:
7 years ago
Why not 3/5 for selecting 3 red balls from total 5 red balls?
(1)
Akanksha said:
6 years ago
A box contains 6000 cards numbered 1 to 6000. One card is drawn at random from the box. Find the probability that the number is neither divisible by 4 nor 6? Please give me a solution.
(1)
Sinchana said:
6 years ago
If the answer is 0.02 something how we can convert please explain it now.
(1)
Subash said:
6 years ago
Can any one explain clearly.
Pavanmahesh said:
5 years ago
All of them red balls means why it can be a 15c3?
Laxman said:
5 years ago
@USHA.
For example, there are 3 red balls and you have to select two balls out of it then possible ways are,
6 ways let me show;
R1,R2 R1,R3 R2,R1 R2,R3 R3,R1 R3,R1.
In these selections, we have R1, R2 and R2, R1 same like the way we consider only one so that we have 3 choices but don't consider you are selecting 3 out of 3 balls.
For example, there are 3 red balls and you have to select two balls out of it then possible ways are,
6 ways let me show;
R1,R2 R1,R3 R2,R1 R2,R3 R3,R1 R3,R1.
In these selections, we have R1, R2 and R2, R1 same like the way we consider only one so that we have 3 choices but don't consider you are selecting 3 out of 3 balls.
Deki Zangmo said:
5 years ago
Total ball = 15.
Out of these, you have to select three balls.
15C3= 455,. We can select 3 balls in 455 ways.
But there is a condition that all 3 balls should be red so,
Out of 5 red balls we have to take out 3;
5C3 = 10,
As a result:
10/455 = 2/91.
Out of these, you have to select three balls.
15C3= 455,. We can select 3 balls in 455 ways.
But there is a condition that all 3 balls should be red so,
Out of 5 red balls we have to take out 3;
5C3 = 10,
As a result:
10/455 = 2/91.
(3)
Poojitha said:
4 years ago
Total balls are 15.
Only we have to select red balls in total balls red are 5.
One selected from a total 5/13.
Next, another selected (we have already selected one so 1 ball is reduced from both red balls and total balls) == 4/14.
Here, 2 balls are reduced 1 from one selected ball another from again selected ball so 3/13
5/15 * 4/14* 3/13 = 2/91.
Only we have to select red balls in total balls red are 5.
One selected from a total 5/13.
Next, another selected (we have already selected one so 1 ball is reduced from both red balls and total balls) == 4/14.
Here, 2 balls are reduced 1 from one selected ball another from again selected ball so 3/13
5/15 * 4/14* 3/13 = 2/91.
(1)
Jino said:
4 years ago
@Deki Zangmo.
Thanks for the clear and best explanation.
Thanks for the clear and best explanation.
(1)
SUDARSAN SWAIN said:
2 years ago
Why 5c2 is being calculated instead of 5c3? Please explain me.
(13)
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