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Ira Lisetskai [31]
3 years ago
12

Twenty one-slips of paper are each marked with a different letter of the alphabet and placed in a basket. A slip is pulled out,

it’s letter recorded and the slip is replaced. This is done 6 times. Find the probability that the word riddle is formed. Assume that each letter in the word is also in the basket

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
6 0

Answer:

4 / 85766121  or 0.00000004664  to  11 decimal places.

or about 5 in 1 hundred million!

Step-by-step explanation:

Probability ( any letter other than d is drawn) = 1/21 as only d occurs more than once in riddle.

Probability (a letter d is drawn) = 2/21  because there are 2 d's in riddle.

To get our answer we multiply probabilities because each drawing of a letter is independent of other draws.

So the Probability of the word drawn being riddle

= 1/21* 1/21*2/21*2/21*1/21*1/21.

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<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units</h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and      y_3=8

The area of a triangle is= \frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]

=|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|

=|-54| = 54 square units

The length of AC = \sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}

                          = \sqrt{(2-12)^{2} +(1-8)^2}

                         =\sqrt{149} units

Let the perpendicular distance from B to AC be = x

According To Problem

\frac{1}{2} \times  x \times \sqrt{149} = 54

⇔x =\frac{108}{\sqrt{149} } units

Therefore the perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units

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