Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell

Question

Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell without replacing any of them, what is the probability that Tarrell will get both a 2 and a 3?

A. 1/5 
B. 1/45 
C. 1/50 
D. 1/90 
E. 14/45

Archit3110 · Answer

total cards ; 10
so getting 2 and 3 ; 2/10 * 1/9 ; 1/45
IMO B

philipssonicare · Answer

Using a combinations approach

If you're unfamiliar with combinations notation: https://www.mathsisfun.com/combinatoric ... tions.html

There is only one '2' to choose and only one '3' to choose (the numerator), but there are 10 numbers from which we need to choose 2 (the probability space)

\(\frac{(1C1·1C1)}{10C2}=\) \(\frac{1·1}{\frac{10!}{8!2!}}=\) \(\frac{1}{\frac{10·9}{2}}=\) \(\frac{1}{45}\)

Can also recognise that the initial deal has a \(\frac{2}{10}\) chance of being a 2 or 3 and that the second deal has a \(\frac{1}{9}\) chance of being the other number
\(\frac{2}{10}·\frac{1}{9}=\frac{2}{90}=\frac{1}{45}\)

ScottTargetTestPrep · Answer

Solution:

The probability Tarrell will get a 2 followed by a 3 is 1/10 x 1/9 = 1/90. Similarly, the probability Tarrell will get a 3 followed by a 2 is also 1/10 x 1/9 = 1/90. Therefore, the probability that Tarrell will get both a 2 and a 3 (in either order) is 1/90 + 1/90 = 1/45.

Alternate Solution:

There are 10C2 = 10!/(2!*8!) = (10 x 9)/2! = 45 ways to pick two cards out of ten cards. A selection of 2 and 3 is one of these 45 ways; therefore, the probability that Tarrell will get both a 2 and a 3 is 1/45.

Answer: B

CEdward · Answer

Ahh the classic trap choice.

Here's my mistake:

1/10 x 1/19 = 1/90

But, wait! I forgot the order...you can get either a 2 first or a 3 first ---> 2,3 OR 3,2

1/90 x 2 = 1/45

Answer is B.

kritisachan · Answer

probability that the two cards will be 2 "and" 3

total cards 10, giving two cards

first card; can be either 2 or 3= 2/10
second card: 1/9

2/10 * 1/9 = 2/90 = 1/45
option B.

mysterymanrog · Answer

We want both a 2 and 3. There are (2!=2) ways of doing this. 
P(2,3)=(1/10)(1/9)=1/90
P(3,2)=(1/10)(1/9)=1/90
2/90 is the same as 1/45, which is B.

luisdicampo · Answer

Deconstructing the Question

Method: Combinations

1. Total Outcomes (Denominator)  
Tammie deals 2 cards out of 10 without replacement. The order in which Tarrell receives the cards does not change the resulting hand.
 Total = 10C2 = (10 * 9) / 2 =    45

2. Favorable Outcomes (Numerator)  
There is only  1  specific combination that contains both a 2 and a 3:  {2, 3} .

3. Probability  
 P = Favorable / Total 
 P =    1 / 45

Alternative Method (Sequential): 
P(2 then 3) + P(3 then 2)
 = (1/10 * 1/9) + (1/10 * 1/9) = 2/90 = 1/45

The correct answer is B.

Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell

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Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell

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