A box contains 15 red marbles, 15 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that atleast one will be green?a)b)c)zerod)1Correct answer is option 'A'. Can you explain this answer?

Question

Answer

It's like doing the question orally...  (don't get into maths... as I was also stuck in the middle)  but by common sense (which is very uncommon)  it can be said as 1-probablity of not getting any green in 5 balls...  in this case we can see the answer aproaches to 1 which can be seen in option a) ....  hope it helps...  but with maths a can see a nightmare coming...

Answer

Sir I cannot explain it

Answer

Calculation:
The total number of marbles in the box is 15 + 15 + 30 = 60.

Finding the probability:
To find the probability of drawing at least one green marble, we need to calculate the probability of the complement event (i.e., the event of not drawing any green marbles) and subtract it from 1.

Finding the probability of not drawing any green marbles:
The number of ways to choose 5 marbles out of the 60 marbles in the box is given by the binomial coefficient C(60, 5).

The number of ways to choose 5 marbles without any green marbles is given by the binomial coefficient C(45, 5) because we are only choosing from the 15 red marbles and 15 blue marbles.

Therefore, the probability of not drawing any green marbles is C(45, 5) / C(60, 5).

Finding the probability of drawing at least one green marble:
Now, we can calculate the probability of drawing at least one green marble by subtracting the probability of not drawing any green marbles from 1.

Probability of drawing at least one green marble = 1 - (C(45, 5) / C(60, 5))

Simplifying the expression:
C(45, 5) = (45*44*43*42*41) / (5*4*3*2*1)
C(60, 5) = (60*59*58*57*56) / (5*4*3*2*1)

Substituting these values into the expression:
Probability of drawing at least one green marble = 1 - [(45*44*43*42*41) / (60*59*58*57*56)]

Calculating the expression gives us the answer as 4367/4484.

Therefore, the probability that at least one marble drawn will be green is 4367/4484, which is option A.

Answer

Probability of getting atleast one green = 1- probability of getting no green = 1-(30C5/60C5) = 4367/4484

Answer

Problem:
A box contains 15 red marbles, 15 blue marbles, and 30 green marbles. 5 marbles are drawn from the box. What is the probability that at least one will be green?

Solution:
To calculate the probability that at least one marble will be green, we can calculate the probability that no marbles are green and subtract it from 1.

Calculating the probability that no marbles are green:
To calculate the probability that no marbles are green, we need to consider the total number of marbles and the number of non-green marbles.

Total number of marbles = 15 + 15 + 30 = 60
Number of non-green marbles = 15 + 15 = 30

We need to choose 5 marbles from the 30 non-green marbles. This can be done in C(30, 5) ways.

C(n, r) = n! / (r! * (n-r)!)

C(30, 5) = 30! / (5! * (30-5)!)

Simplifying this expression gives us C(30, 5) = 142506.

Therefore, the probability of choosing 5 non-green marbles is 142506 / C(60, 5).

Calculating the probability of choosing at least one green marble:
The probability of choosing at least one green marble = 1 - probability of choosing 5 non-green marbles.

So, the probability of choosing at least one green marble = 1 - (142506 / C(60, 5)).

To calculate C(60, 5), we can use the formula C(n, r) = n! / (r! * (n-r)!).

C(60, 5) = 60! / (5! * (60-5)!)

Simplifying this expression gives us C(60, 5) = 5461512.

Therefore, the probability of choosing at least one green marble = 1 - (142506 / 5461512) = 4367/4484.

Hence, the correct answer is option A) 4367/4484.

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