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Alexandra [31]
3 years ago
8

If one card is randomly chosen from a standard deck, what is the probability that a 6 is chosen, GIVEN THAT A DIAMOND WAS CHOSEN

?
A. 4/13
B. 1/52
C. 1/13
D. 1/4
Mathematics
1 answer:
Paraphin [41]3 years ago
3 0

Normally I would use the formula for conditional probability, but I think it would be overdoing.


Since we know the chosen card is a diamond, then there are 13 possible cards. There is only one 6 of diamonds, so the probability is \dfrac{1}{13}

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An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel.
Greeley [361]

Answer: 1) 0.6561    2) 0.0037

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

P(X=x)=^nC_xp^x(1-p)^{n-x}

, where p =Probability of getting success in each trial.

As per given , we have

The probability that any satellite dish owners subscribe to at least one premium movie channel.  : p=0.10

Sample size : n= 4

Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.

1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}

=(1)(0.90)^4=0.6561

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.

2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

= P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.

8 0
3 years ago
How do you find the answer
kati45 [8]
The answer is (2, 0).
4 0
3 years ago
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