Question

Annie writes the numbers 1 through 10 on note cards. She flips the cards over so she cannot see the number and selects three cards from the stack. What is the probability that she has selected the cards numbered 1, 2, and 3?

Answer 1

Answer:

Probability that she has selected the cards numbered 1, 2, and 3 is  $\frac{1}{720}$ .

Step-by-step explanation:

We are given that Annie writes the numbers 1 through 10 on note cards. She flips the cards over so she cannot see the number and selects three cards from the stack.

Here, the total number of cards = 10 and the cards are numbered as 1, 2, 3 ,4, ...., 10.

Probability of any event = Favorable outcomes ÷ Total no. of outcomes

Probability of picking a card with number 1 = (1/10)

Now after the first card is picked, the number of cards left = 10 - 1 = 9

So, Probability of picking a card with number 2 = (1/9)

Similarly, Probability of picking a card with number 3 = (1/8)

The total probability that she has selected card with number 1, 2 and 3 is given by;

            (1/10) * (1/9) * (1/8) = (1/720)

Therefore, the probability that she has selected the cards numbered 1, 2, and 3 is (1/720) .

Answer 2

The probability that Annie has selected the cards numbered 1, 2, and 3 is $(\frac{1}{720} )$

Step-by-step explanation:

Here, the total number of cards in the set = 10

The cards are numbered as 1,2,3,4,5,...., 10

P(Any event E)  = $\frac{\textrm{Total favorable events}}{\textrm{Total number of events}}$

P(picking a card with number 1)  = $\frac{\textrm{Total cards with number 1 on it}}{\textrm{Total cards}} = (\frac{1}{10})$

Now when the first card is picked, the number of cards left = 10 - 1 = 9

P(picking a card with number 2)  = $\frac{\textrm{Total cards with number 2 on it}}{\textrm{Total cards}} = (\frac{1}{9})$

Similarly, P(picking a card with number 3)  = $\frac{\textrm{Total cards with number 3 on it}}{\textrm{Total cards}} = (\frac{1}{8})$

So, the total probability  that she has selected card with number 1, 2 and 3

= $(\frac{1}{10} ) \times (\frac{1}{9} ) \times (\frac{1}{8} ) = (\frac{1}{720} )$

Hence,  the probability that she has selected the cards numbered 1, 2, and 3 is $(\frac{1}{720} )$