The diagram the area of the large square is one square unit 2000 new segments divide a square into 4 equal sized triangles two o
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The
formula , where r! is 1*2*3*...r
is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.
for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.
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The bucket contains 6 orange balls and 7 yellow balls, that is a total of 13 balls.
The total number of groups of 5 balls out of the 13, that we can form is :
<span>1) What is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow?
the event can happen as follows:
{1 orange, 4 yellow} or </span>{2 orange, 3 yellow} or {3 orange, 2 yellow} or {4 orange, 1 yellow}
that is, the only cases we are not considering are {5 orange} or {5 yellow}
5 orange balls can be selected in C(6, 5) = 6!/(5!1!)=6 different ways,
and 5 yellow balls can be selected in C(7, 5)=7!/(5!2!)=(7*6)/2=21 different ways.
so 5 orange balls or 5 yellow balls can be selected in 6+21=27 different ways,
this means that P(5 orange balls or 5 yellow balls)=27/1287=0.02
P({1 orange, 4 yellow} or {2 orange, 3 yellow} or {3 orange, 2 yellow} or {4 orange, 1 yellow}) = 1-P(5 orange balls or 5 yellow balls)=1-0.02=0.98
Part 2
The event "at least two are orange and at least two are yellow " can happen in the following ways:
{2 orange, 3 yellow} or {3 orange, 2 yellow}
2 orange, 3 yellow balls can be selected in C(6,2)*C(7,3) because
any selected 2 of the 6 orange balls, can be combined with any group of 3, out of the 7 yellow balls.
so we calculate:
thus,
P ({2 orange, 3 yellow} or {3 orange, 2 yellow})=
P (2 orange, 3 yellow) + P (3 orange, 2 yellow)=
The given graph is a Distance- Time graph whose slope gives us speed that is ~ rate of Distance covered ( m ) and per unit time ( sec ).
therefore, the correct choice is ~