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azamat
4 years ago
14

Ten slips of paper numberd 1 through 10 are placed in a bag. You draw a slip at random and draw another without replacing the fi

rst.Find the probabilty that both numbers are odd
Mathematics
1 answer:
Bogdan [553]4 years ago
4 0

Answer:

Probability that both numbers are odd =2/9

Step-by-step explanation:

Odd numbers are numbers such that, when they divide 2, there is a remainder.

Ten slips of paper numbered 1 through 10 are placed in a bag.

In number one to ten, the odd numbers are 1, 3,5,7 and 9

They all divide 2 and leaves a remainder.

Total number of odd numbers is 5

Total number of both odd and even numbers from one through 10 is 10

Probability of picking an odd number

= number of odd numbers/ total numbers.

Probability of picking an odd number

= 5/10= 1/2

Probability of picking another odd number without replacement is 4/9

Probability that both numbers are odd = 1/2 × 4/9 =2/9

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Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

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Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

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                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

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