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sp2606 [1]
3 years ago
7

After removing all of the clubs from a deck of cards, you are left with a 39 card deck with Hearts, Diamonds, and Spades. Answer

the following questions assuming that after each draw of a card, that card is returned to this deck and reshuffled.
What is the probability of :A) drawing a red card ?B) drawing a heart or a red card?C) drawing a jack or a red card?
Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer:

(a)2/3

(b)2/3

(c)9/13

Step-by-step explanation:

Total Number of Cards in new Deck=39

Hearts(Red)=13

Diamonds(Red)=13

Spades(Black)=13

(a)P(drawing a red card)

Total number of red cards = 13+13=26

P(drawing a red card)=26/39=2/3

(b)Drawing a heart or a red card

Number of Hearts=13

Number of red cards=26

Number of Red Hearts = 13

Since the two events are not mutually exclusive

P(Hearts or Red) = P(Hearts) + P(Red) - P( Hearts and Red)

P(H∪R)=P(H)+P(R)-P(H∩R)

=13/39 + 26/39 - 13/39

=26/39 =2/3

(c)Drawing a jack or a red card.

Number of Jacks=3

Number of red cards=26

Number of Red Jacks = 2

Since the two events are not mutually exclusive

P(Jack or Red) = P(Jacks) + P(Red) - P( Jacks and Red)

P(J∪R)=P(J)+P(R)-P(J∩R)

=3/39 + 26/39 - 2/39

=27/39 =9/13

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Step-by-step explanation:

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3 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

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where, X bar = sample mean weight

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P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

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3 years ago
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