a jar contains 6 jellybeans, 4 green jellybeans, and 4 blue jelly beans. if we choose a jellybean, then another without putting
the first one back in the jar, what is the probability that the first jelly bean will be blue, and the second one will be blue as well?
1 answer:
Probability first one is blue is 4/6+4+4=4/14
the second one is also blue is 4-1/6+4+4-1=3/13
The probability of getting blue both time is 4/14*3/13=12/182=0.0659=6.59%
Answer:
0.0659 or 6.59%
You might be interested in
Answer:
740000
Step-by-step explanation:
100ps make up £1, so times 7400 by 100 to get the answer
Answer:
i think that would be (2,0)
Answer:
none of the answer above
Step-by-step explanation:
it should be
16x +8 or 8(2x+y)
group the number together according to alphabet and number
Answer:
Step-by-step explanation:
We know:
We have
Use 
![\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2%2B%5Ccos%5E2%5Ctheta%3D1%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B4%7D%2B%5Ccos%5E2%5Ctheta%3D1%5Cqquad%5Ctext%7Bsubtract%7D%5C%20%5Cdfrac%7B1%7D%7B4%7D%5C%20%5Ctext%7Bfrom%20both%20sides%7D%5C%5C%5C%5C%5Ccos%5E2%5Ctheta%3D%5Cdfrac%7B4%7D%7B4%7D-%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5C%5Ccos%5E2%5Ctheta%3D%5Cdfrac%7B3%7D%7B4%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Csqrt%7B%5Cdfrac%7B3%7D%7B4%7D%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Cdfrac%7B%5Csqrt3%7D%7B%5Csqrt4%7D%5Cto%5Ccos%5Ctheta%3D%5Cpm%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%5C%5C%5C%5C%5Ctheta%5Cin%5B0%5Eo%2C%5C%2090%5Eo%5D%2C%5C%20%5Ctext%7Btherefore%20all%20functions%20have%20positive%20values%20or%20equal%200.%7D%5C%5C%5C%5C%5Ccos%5Ctheta%3D%5Cdfrac%7B%5Csqrt3%7D%7B2%7D)
