9. How many different 10 letter words (real or imaginary) can be formed from the

Question

9. How many different 10 letter words (real or imaginary) can be formed from the following letters?

X,C,F,B,H,Y,C,T,O,Q

__?__ ten-letter words (real or imaginary) can be formed with the given letters

(Type a whole number.)

1. Suppose that E and F are two events and that P(E and F)= 0.6 and P(E)=0.8. What is P(F|E) ?

P(F|E)= __?___ (Type an integer or a decimal.)

4. (a) What is the probability that two randomly selected tulip bulbs are both red?

(Round to three decimal places as needed.)

(b) What is the probability that the first bulb selected is red and the second yellow?

(Round to three decimal places as needed.)

(c) What is the probability that the first bulb selected is yellow and the second red?

(Round to three decimal places as needed.)

(d) What is the probability that one bulb is red and the other yellow?

(Round to three decimal places as needed.)

Explanation / Answer

Ans:

9)

C is repeating twice,so

Number of possible 10-letter word=10!/2=1814400

(10!=10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800)

1)P(E and F)= 0.6 and P(E)=0.8

P(F/E)=P(E and F)/P(E)=0.6/0.8=3/4=0.75

4)A bag of 29 tulips contains 14 red, 8 yellow, and 7 purple.

  1) Probability first bulb is red = 14/29
There are not 13 red bulbs left and 28 total bulbs left
Probability second bulb is red = 13/28
p = 14/29 * 13/28 = 13/58 = 0.224

2) Probability first bulb is red = 14/29
Probability second bulb is yellow = 8/28

p = 14/29 * 8/28 = 4/29 = 0.138

3)
Probability first bulb is yellow = 8/29
Probability second bulb is red = 14/28

p = 8/29 * 14/28 = 4/29 = 0.138

4)
p = 4/29 + 4/29 = 8/29 = 0.276

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