What is the probability that a randomly drawn hand of four cards contains all black cards or all face cards? The probability is 6 Round to four decimal places as needed.)


Answer 1

Answer: 0.05699

Step-by-step explanation:

The total number of cards in a deck = 52

The total number of black cards = 26

Then ,\text{P(Black)}=(C(26,4))/(C(52,4))=0.00182842367\approx0.00183

The total number of face cards = 12

Then , \text{P(Face)}=(C(12,4))/(C(52,4))\approx0.05522

The number of cards that are black and face cards = 6

Then , \text{P(Black and Face )}=(C(6,4))/(C(52,4))\approx0.00006

Then , the probability that a randomly drawn hand of four cards contains all black cards or all face cards is given by :-

\text{P(Black or Face)}=\text{P(Black)+P(Face)-P(Black and Face)}\n\n\Rightarrow\ \text{P(Black or Face)}=0.00183+0.05522-0.00006\n\n\Rightarrow\ \text{P(Black or Face)}=0.05699

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The answer I that u are looking for is 13

Find the 12th term in the sequence: 1/16, -1/8, 1/4



Step-by-step explanation:

hello :

the sequence is geometric because : (1/4)/(-1/8) = (-1/8)/(1/16)=-2   = r

the nth term is :  An =A1 × r^(n-1)

a common ratio is : r  and A1 the first term

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