[tex] \displaystyle
|\Omega|=\binom{6}{2}=\dfrac{6!}{2!4!}=\dfrac{5\cdot6}{2}=15\\
|A|=\underbrace{1\cdot5}_{\text{1 and X}}+\underbrace{1\cdot5}_{\text{4 and X}}-\underbrace{1}_{\text{1 and 4 }} [/tex]
[tex] |A|=5+5-1=9\\\\
P(A)=\dfrac{9}{15}=\dfrac{3}{5}=60\% [/tex]
Answer:
3/5
Step-by-step explanation:
There are 6 possibilities for the number on the first card Mary draws. After this, there are 5 possibilities for the second card, because Mary can't draw the first card again. Thus, there are 6(5)=30 possibilities for the pair of cards.
Among these, there are 4(3) = 12 pairs that don't include at least one square number. To see why, consider that there are 4 choices for the first non-square number, then 3 choices for the second non-square number (since $1$ of the original $4$ choices has been used up).
Since 12 possible pairs don't include at least one square, the other 30-12=18 pairs do include at least one square. Therefore, the probability that Mary gets at least one square number is 18/30=3/5.
