Pair Almond
Pair Almond
probability?
1-a group of volleyball players consists of four grade 11 students and six grade 12 students. if six players are chosen at random to start a match, whatis the probability that three will be from each grade?
2-if a bowl contains ten hazelnuts and eight almonds, what is the probability that four nut randomly selected from the bowl will all be hazelnuts?
3-without looking Jenny randomly selects two socks from a drawer containing four blue, three white, and five black socks, none of which are paired up. what is the probability that she chooses two socks of the same colour?
4-a euchre deck has 24 cards; the 9, 10, jack, queen, king and the ace of each suit. if you were to deal out five cards from this deck, what is the probability that they will be a 10, jack, queen, king, and ace all from the same suit?
You can use the hypergeometric distribution to find the solution
Let X be the number of objects of a give type drawn. X has the hypergeometric distribution with the following parameters.
K = number of items to be drawn
N = total objects
M = number of objects of a given type
The probability mass function for the hypergeometric distribution is defined as:
P(X = x | N, M, K) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K ) for x = {0, ..., K}
P(X = 0 | N, M, K) = 0 otherwise
Note that the constraints on x here are very generic and it is possible to have value of K, N and M such that for x in {0, ..., K} P(X = x) = 0.
If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr
the N C K is the total number of possible combinations of K objects drawn from N objects.
the M C x is the number of combinations of getting x objects of the given type
the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn.
Looking at the PMF you should be able to see that it is the ratio of the number of combination of selecting the X of the items of interest times the number of combinations of choosing K - X items from the remaining items and this is all divided by the total number of combination for choosing K items from N objects.
===
for question one we have:
X = number of 11 grade players
K = number of items to be drawn = 6
N = total objects = 10
M = number of objects of a given type = 4
P(X = 3 ) = 0.3809524
this is the prob that 3 of the players are from grade 11 and the other three are from grade 12.
===
2)
X = number of hazelnuts
K = number of items to be drawn = 4
N = total objects = 18
M = number of objects of a given type = 10
P(X = 0 ) = 0.02287582
P(X = 1 ) = 0.1830065
P(X = 2 ) = 0.4117647
P(X = 3 ) = 0.3137255
P(X = 4 ) = 0.06862745
===
3)
P(2 blue) = 4/12 * 3/11
P(2 white) = 3/12 * 2/11
P(2 black) = 5/12 * 4/12
P(pair of same colour)
= P(2blue) + P(2white) + P(2black)
= 38/132 = 0.2878788
===
4)
If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr
there are 24 C 5 = 42504 possible hands
there are (4 C 1) ^ 5 = 1024 possible hands with a 10, jack, queen, king and ace
probability is 1024 / 42504 = 0.02409185
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