Acoin is tossed 4 times. what is the probability if getting 2 heads, 2 tails?

11/16

Step-by-step explanation:

P(THHH)=4!3!∗(12)4=416P

(THHH)=4!3!∗(12)4=416, we are multiplying by 4!/3! since THHH scenario can occur in number of ways: THHH, HTHH , HHTH, or HHHT (notice that 4!/3! basically gives number of arrangements of 4 letters THHH out of which 3 H's are identcal).

P(T≥2)=1−(1/16+4/16)=11/16

the answer is a

step-by-step explanation:

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: