In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are
possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by
. So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability
![\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cbinom%7B13%7D1%5Cbinom44%5Cbinom%7B48%7D1%7D%7B%5Cbinom%7B52%7D5%7D%3D%5Cdfrac%7B624%7D%7B2%2C598%2C960%7D%5Capprox0.00024)
There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing
possible hands. Exactly 2 aces are drawn in
hands. And so on. This gives a total of
![\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Ba%3D1%7D%5E4%5Cbinom4a%5Cbinom%7B48%7D%7B5-a%7D%3D886%2C656)
possible hands containing at least 1 ace, and hence B occurs with probability
![\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csum%5Climits_%7Ba%3D1%7D%5E4%5Cbinom4a%5Cbinom%7B48%7D%7B5-a%7D%7D%7B%5Cbinom%7B52%7D5%7D%3D%5Cdfrac%7B18%2C472%7D%7B54%2C145%7D%5Capprox0.3412)
The product of these probability is approximately 0.000082.
A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e.
. This happens if
- the hand has 4 aces and 1 non-ace, or
- the hand has a non-ace 4-of-a-kind and 1 ace
The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So
consists of 96 possible hands, which occurs with probability
![\dfrac{96}{\binom{52}5}\approx0.0000369](https://tex.z-dn.net/?f=%5Cdfrac%7B96%7D%7B%5Cbinom%7B52%7D5%7D%5Capprox0.0000369)
and so the events A and B are NOT independent.
Answer:
3?
Step-by-step explanation:
i think that because it's multiplying by 3.
3*3= 9
5*3= 15
7*3= 21
9*3= 27
This is the only way i could think of helping in some sort of way! I'm sorry if it's wrong! (´。_。`)
Have a great day! (´。_。`) / (ノ◕ヮ◕)ノ*:・゚✧
( ゚д゚)つ Bye
3.6 ounces --> 5.9 ounces
4.2 ounces --> 6.5 ounces
3.3 ounces --> 5.6 ounces
You just basically add 2.3 ounces to each of the kitten's original weight.