a bag contains 6 red marbles, 5 white marbles, and 5 blue marbles. you draw 4 marbles out at random, without…

a bag contains 6 red marbles, 5 white marbles, and 5 blue marbles. you draw 4 marbles out at random, without replacement. what is the probability that all the marbles are red?\nthe probability that all the marbles are red is \nwhat is the probability that exactly two of the marbles are red?\nthe probability that exactly two of the marbles are red is \nwhat is the probability that none of the marbles are red?\nthe probability of picking no red marbles is

a bag contains 6 red marbles, 5 white marbles, and 5 blue marbles. you draw 4 marbles out at random, without replacement. what is the probability that all the marbles are red?\nthe probability that all the marbles are red is \nwhat is the probability that exactly two of the marbles are red?\nthe probability that exactly two of the marbles are red is \nwhat is the probability that none of the marbles are red?\nthe probability of picking no red marbles is

Answer

Explanation:

Step1: Calculate total number of marbles

The total number of marbles is $6 + 5+5=16$ marbles.

Step2: Calculate probability that all 4 marbles are red

The number of ways to choose 4 red marbles out of 6 red marbles is $C(6,4)=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15$. The number of ways to choose 4 marbles out of 16 marbles is $C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820$. The probability $P_1=\frac{C(6,4)}{C(16,4)}=\frac{15}{1820}=\frac{3}{364}$.

Step3: Calculate probability that exactly 2 marbles are red

The number of ways to choose 2 red marbles out of 6 is $C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$. The number of ways to choose the remaining $4 - 2 = 2$ non - red marbles out of $16-6 = 10$ non - red marbles is $C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45$. The number of ways to choose 4 marbles out of 16 is $C(16,4) = 1820$. The probability $P_2=\frac{C(6,2)\times C(10,2)}{C(16,4)}=\frac{15\times45}{1820}=\frac{675}{1820}=\frac{135}{364}$.

Step4: Calculate probability that none of the marbles are red

The number of ways to choose 4 non - red marbles out of 10 non - red marbles is $C(10,4)=\frac{10!}{4!(10 - 4)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$. The number of ways to choose 4 marbles out of 16 is $C(16,4)=1820$. The probability $P_3=\frac{C(10,4)}{C(16,4)}=\frac{210}{1820}=\frac{3}{26}$.

Answer:

The probability that all the marbles are red is $\frac{3}{364}$. The probability that exactly two of the marbles are red is $\frac{135}{364}$. The probability of picking no red marbles is $\frac{3}{26}$.