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oee [108]
3 years ago
10

daniel has 6 different pants that match with 4 different shirts how many shirt and pants combinations can he make if he selects

one shirt and one pair of pants?
Mathematics
2 answers:
igor_vitrenko [27]3 years ago
6 0
So, for each shirt, there are six possible pants combinations. Because there are 4 shirts, there are 4 * 6 = 12 total combinations
Sholpan [36]3 years ago
5 0
6 pants times 4 shirts is 24 different combinations (not twelve, previous answerer)
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If you are dealt 3 cards from a shuffled deck of 52 cards , find the probability that all 3 cards are clubs .
polet [3.4K]
Probability of having a club:

1st card = 13/52
2nd card= 12/51
3rd card = 11/50

Total conditional probability = 13/52 * 12/51 * 11/50= 0.0129 =1.29%

4 0
3 years ago
Two six-sided dice are rolled (one red and one green). Some possibilities are (Red=1,Green=5) or (Red=2,Green=2) etc.
klasskru [66]
A.) There are going to be 72 total possibilities. 
Because if you have 6 times 6 is 36 times 2 is 72. 
B.) The probability that the sum on the dice come out to be 11 is 2:72 
Because you can only have (6,5) and (5,6) 
C.) The probability that the sum on the two dice comes out to be 7 is 6:72 
Because you have (6,1) (1,6) (5,2) (2,5) (4,3) (3,4) 
D.) The probability that the numbers on the two dice are equal is 6:72 
<span>Because you have (6,6) (5,5) (4,4) (3,3) (2,2,) (1,1) </span>
6 0
4 years ago
Find (12y-1) + (-10y+7)
dezoksy [38]

Answer:

2y +6

Step-by-step explanation:

(12y-1) + (-10y+7)

We need to combine like terms

12y - 10y  + -1 +7

2y +6

4 0
3 years ago
Please help me with this
Dafna11 [192]
I hope this helps you.

8 0
4 years ago
An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6
SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6  times.

The above situation can be represented through binomial distribution;

P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......

where, n = number trials (samples) taken = 6 trials

            r = number of success = at least 4

           p = probability of success which in our question is probability of

                 rolling a “six", i.e; p = 0.20

<u><em>Let X = Number of sixes on a die</em></u>

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X \geq 4)

P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)

=  \binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}

=  15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}

=  0.0154 + 0.00154 + 0.000064

=  0.01696

<em />

Therefore, probability of rolling at least 4 sixes is 0.01696.

8 0
3 years ago
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