Probability Test - 5

Again, we are not given any absolute lengths, so we will pick something convenient. Here, I am going to pick WZ = XZ = 1.  Triangle WXZ is an  Isosceles Right Triangle, i.e. a 45-45-90 triangle. Its area is A= (1/2)bh, and since both base and height are 1, the area of the triangle is 1/2.
Both WX and WY are radii of the circle. From the proportions in the 45-45-90 triangle, we know

Now, we can find the area of the circle, using Archimedes ’formula:

Because triangle WXZ is a 45-45-90 triangle, we know ∠W = 45 °.  Well, a 45 °angle must be half of 90 °—- which means a 45 °is 1/4 of 180 °—- which means a 45 °is 1/8 of 360 °. Thus, a 45 °sector would be 1/8 of a circle and have 1/8 of the entire circle ’s area.

First of all, that ’s the denominator-area in our probability. Also, we can use this to figure out the area of the shaded region.

Area of shaded region = (area of sector) –(area of triangle)

That ’s our numerator-area. Now, we can put together the probability —we will just have to simply the complex fraction by multiplying numerator and denominator by 4.

Here, we are not given a diagram, so we will sketch one. In the real test, just a rough diagram will be enough to visualize things. Here ’s an accurate diagram.

Notice the square JKLM has an area of 6 x 6 = 36 —- right there, that ’s our denominator-area.

Now, let ’s think about this line.  Let ’s solve for y, to put the line into slope-intercept form (i.e. y = mx + b form).

First of all, putting it in slope-intercept form makes clear —the y-intercept of this line is point S, (0, –2), which is on one side of the square.  The slope is 3/5, so over five and up three puts us at point T, (5, 1), also on a side of the square.  Triangle STM has a base of 5 and height of three, so the area is

That ’s the numerator-area, so just divide

We aren ’t given any absolute lengths.  For convenience, I am going to assume that the radius of the circle is r = 1.  That ’s very easy.  Right away, we know the area of the circle is pi .  Notice, the height of the triangle is equal to the diameter of the circle, so h = AB = 2.  We are told the ratio of AB : BC = 1 : 2, so BC, the width, must equal w = BC = 4.  Area of the rectangle is h*w = (AB)*(BC) = 8.  That, right there, is our “denominator area ”.  Now, for the numerator area, the area of the rectangle that does not include the circle, subtract the circle from the rectangle: A = 8 –pi .  That ’s our numerator.

For the denominator, we are going to pick two books from among ten total: a combination of two from ten. Again, we will use  the formula:

which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n –1) positive integers. Here

That ’s the total number of pairs of books we could pick from the ten on the shelf. That ’s our denominator.

Now, the numerator. We want one novel and one reference work. Well, there are four novels and two reference works, so by  the FCP, the number of ways we can pick this is 4 x 2 = 8. That ’s the total possible number of pairs involving exactly one of these four novels and exactly one of these two reference works. That ’s our numerator.

First, we will count all the possible arrangements of the five children on the five seats, all the possible orders.  This is 5! = 120. That ’s the denominator.

Now, the more challenging part: we have to figure out how many arrangements there are involving C &E sitting together. This is a tricky problem to frame, so I ’ll demonstrate the steps to follow. First, let ’s look at the seats these two could be next to each other.  There are four possible pairs of seats in which they could be next to each other

i. X X _ _ _

ii. _ X X _ _

iii. _ _ X X _

iv. _ _ _ X X

In each of those four cases, we could have either CE or EC, either order, so that ’s 4 x 2 = 8 ways we could have just C &E sitting next to each other with the remaining three seats empty.

For the final step, we need to consider the other three children, A &B &D. In each of the eight cases, there are three blank seats waiting for those three, and those three could be put in any order in those blank seats. Three elements in any order —that ’s 3! = 6. Thus, the total number of arrangements in which C &E would be next to each other would be 8 x 6 = 48. This is our numerator.

The probability would be this number, 48, over the total number of arrangements of the children, 120.

A full deck of 52 cards contains 13 cards from each of the four suits.  The probability of drawing a heart from a full deck is 1/4.  Therefore, the probability of “not heart ”is 3/4.

P(at least three draws to win) = 1 –P(win in two or fewer draws)

Furthermore,

P(win in two or fewer draws) = P(win in one draw OR win in two draws)

= P(win in one draw) + P(win in two draws)

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart.  Above, we already said: the probability of this is 1/4.

P(win in one draw) = 1/4

Winning in two draws means: my first draw is “not heart ”, P = 3/4, AND the second draw is a heart, P = 1/4.  Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win in two draws) =(3/4)*(1/4) = 3/16

P(win in two or fewer draws) = P(win in one draw) + P(win in two draws)

= 1/4 + 3/16 = 7/16

P(at least three draws to win) = 1 –P(win in two or fewer draws)

= 1 –7/16 = 9/16

P(at least one H) = 1 –P(no H ’s)

In one flip, P(“not H ”) = P(T) = 1/2. We would need to have this happen six times —that is to say, six independent events joined by AND, which means they are multiplied together.

P(at least one vowel) = 1 –P(no vowels)

The probability of picking no vowel from the first set is 3/5.  The probability of picking no vowel from the second set is 5/6.  In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set.  According to the AND rule, we multiply those probabilities.

P(no vowels) = (3/5)*(5/6) = 1/2

P(at least one vowel) = 1 –P(no vowels) = 1 –1/2 = 1/2

Solution :- Given letter is  PROBABILITY

So, total letter is 11

no. of vowels  =4

no. of consonant  =7

So  probability of choosing vowel is  4/11

Picking an M is not disjoint with picking a C —they both could happen on the same round of the game.  We have to use the generalized OR rule for this:

P(C or M) = P(C) + P(M) –P(C and M)

Fortunately, we know the first two, and we calculated the value of the third term already in #1.

P(C or M) = P(C) + P(M) –P(C and M)