Question

Mary has six cards whose front sides show the numbers 1,2,3,4,5, and 6. She turns the cards face-down, shuffles the cards until their order is random, then pulls the top two cards off the deck. What is the probability that at least one of those two cards shows a square number?

Answer 1

Answer:

0.6

Step-by-step explanation:

Among numbers 1, 2, 3, 4, 5 and 6, the square numbers are 1 and 4.

The probability that both of these two cards do not show a square number (both selected cards show numbers 2 or 3 or 5 or 6) is

$Pr=\dfrac{C_4^2}{C_6^2}=\dfrac{\frac{4!}{2!(4-2)!}}{\frac{6!}{2!(6-2)!}}=\dfrac{4!\cdot 2!\cdot 4!}{2!\cdot 2!\cdot 6!}=\dfrac{1\cdot 2\cdot 3\cdot 4}{1\cdot 2\cdot 5\cdot 6}=\dfrac{2}{5}.$

The probability that at least one of those two cards shows a square number is

$1-Pr=1-\dfrac{2}{5}=\dfrac{3}{5}=0.6.$

Answer 2

Answer:

3/5

Step-by-step explanation:

The probability that at least one of those two cards shows a square number is 3/5, or 60\%.

The first step would be to figure the square numbers out of 1,2,3,4,5, and 6. It turns out evident that the only numbers that are squared are 1 and 6.

The keyword given in the problem was that it stated: "at least one of those two cards shows a square number".

To solve the problem, the steps involve subtracting the possibility of not choosing the numbers from the total. The first possible choice of not choosing the squared numbers would be 4/6 (since out of all the six cards, there are two that are square numbers). With one card drawn, the remaining choices would be 3/5.

Multiplying the two then subtracting from one, 1-4/6(3/5), would give 3/5.

Therefore, the probability that at least one of the two cards drawn by Mary is 3/5, or 60%.