1) a bag has 4 red ball's, 5 white ball's and 6 blue ball's. 3 ball's are drawn at random from the bag without putting it back. what is the probability that they are of the same color?
2) 2 cards are chosen at random from a standard deck of 52-card. what is the probability that the 14 card is a heart and the second card is a 10?
+2440 Sir Cphill seems to be exercising his mount.
Here’s the solution for #1
\(\text {There are }\binom{3}{20} \text { = 1140 ways to choose to choose any 3 ball's }\\ \text {There are }\binom{3}{4} \text { = 4 ways to choose to choose 3 red ball's } \\ \text {There are }\binom{3}{5} \text { = 10 ways to choose to choose 3 white ball's } \\ \text {There are }\binom{3}{6} \text { = 20 ways to choose to choose 3 blue ball's } \tiny \text {(This sounds painful)}\\ \)
\(\text {For 3 red ball's the probability is } \hspace{.45cm} \dfrac {4}{1120} \\ \text {For 3 white ball's the probability is }\dfrac {10}{1120} \\ \text {For 3 blue ball's the probability is } \hspace{.25cm} \dfrac {20}{1120} \\ \)
\(\text {The overall probability of 3 of the same color is }\\ \frac {1}{3} \text { the harmonic average of these three probabilities. }\\ \text{ }\\ \hspace{1cm} \dfrac {1}{\dfrac {1120}{4} + \dfrac {1120}{10}+ \dfrac {1120}{30}} = \dfrac {1}{448} \)
+118609 Thanks Ginger..... I think you made a couple of little mistakes :)
1)
a bag has 4 red ball's, 5 white ball's and 6 blue ball's. 3 ball's are drawn at random from the bag without putting it back. what is the probability that they are of the same color?
There are 4+5+6=15 b***s altogether.
What is the probability of drawing 3 red b***s
\(\text{The prob of 3 reds is }\frac{4}{15}\times \frac{3}{14} \times \frac{2}{13}=\frac{24}{2730}\\ \text{The prob of 3 whites is }\frac{5}{15}\times \frac{4}{14} \times \frac{3}{13}=\frac{60}{2730}\\ \text{The prob of 3 blues is }\frac{6}{15}\times \frac{5}{14} \times \frac{4}{13}=\frac{120}{2730}\\ \text{add all these together and you get}\\ \text{P(three the same colour )=}\frac{34}{455}\\\)
Another method:
\(\frac{\binom{4 }{3}\times \binom{5}{3}\times \binom{6}{3}}{\binom{15}{3}}=\frac{4+10+20}{455}=\frac{34}{455}\)
2)
2 cards are chosen at random from a standard deck of 52-card. what is the probability that the 1st card is a heart and the second card is a 10?
Mmm
10 of hearts followed by another 10
\(\frac{1}{52} \times \frac{3}{51} = \frac{3}{52*51}=\frac{3}{2652}\)
or
a heart that is not a 10 followed by a 10
\(\frac{12}{52}\times \frac{4}{51}=\frac{48}{2652}\\ ~\\ \text{P(heart then 10)}=\frac{3+48}{2652}=\frac{51}{2652}=\frac{1}{52}\)
+2440 Wow! I didn’t just tangle this. I royally scrambled it. (I inverted the binomials, too)
I’m glad you are here to unscramble the sacred ball’s of Excalibur.
In your “other method” you multiplied the binomials but added the results. I assume for this method that you add binomials, because you are counting, and for the first method, you multiply the probabilities. Is that a correct assumption?
This is a broad question, but when do you use the harmonic mean to find “overall” probability?
Thank you, Miss Melody.
