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liubo4ka [24]
2 years ago
10

Nine cards are numbered from 1 to 9 and placed in a box. One card is selected at random and not replaced. Another card is random

ly selected. What is the probability of selecting two prime numbers?
Mathematics
2 answers:
gizmo_the_mogwai [7]2 years ago
8 0

Answer:

1/6

Step-by-step explanation:

There are 4 primes. So the probability for the first draw is 4/9. Since the card is not replaced, the second probability is 3/8. 3/8 * 4/9 is 12/72, which simplifies into 1/6.

Alex_Xolod [135]2 years ago
8 0

Answer:

1/6

Step-by-step explanation:

There are 4 possible primes: 2, 3, 5, 7

First card has 4/9 probability of being a prime. Second card has 3/8 probability of being a prime(since 3 prime cards left and 8 total cards left because we do not replace the first card). 4/9 * 3/8 = 1/6.

Another way to do this is using combinatorics. C(9,2) = 36 ways to choose 2 distinct cards, C(4,2) = 6 ways to choose two of the "prime" cards. 6/36 = 1/6.

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You have 16 different cuts of flowers and plan to use seven of them how many different selections of the seven flowers are possi
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We can have 11440 different selections.

<u>SOLUTION: </u>

Given, you have 16 different cuts of flowers  

And you have planned to use seven out of them.  

We have to find how many different selections of the seven flowers are possible.  Now, we have to select 7 items out of 16 available items. So we have to use combinations.

Then, 7 out of 16 combination \rightarrow^{16} \mathrm{c}_{7} \rightarrow \frac{16 !}{(16-7) ! 7 !}

\bold{\text{ Since } ^nC_r=\frac{n !}{(r !(n-r) !)}}

\begin{array}{l}{\rightarrow \frac{16 !}{9 ! \times 7 !}} \\\\ {\rightarrow \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 !}{9 ! \times 7 !}} \\\\ {\rightarrow \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{7 \times 6 \times 13 \times 12 \times 11 \times 10}} \\\\ {\rightarrow \frac{57657600}{5040}} \\\\ {\rightarrow 11440}\end{array}

Hence, we can have 11440 different selections.

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