If you are turning clockwise you will land at south
Solution:
The probability of an event is expressed as
![P(event)=\frac{number\text{ of desired outcome}}{number\text{ of possible outcome}}](https://tex.z-dn.net/?f=P%28event%29%3D%5Cfrac%7Bnumber%5Ctext%7B%20of%20desired%20outcome%7D%7D%7Bnumber%5Ctext%7B%20of%20possible%20outcome%7D%7D)
In a pack of 52 cards, we have
![\begin{gathered} number\text{ of diamonds=13} \\ number\text{ of ace= 1} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20number%5Ctext%7B%20of%20diamonds%3D13%7D%20%5C%5C%20number%5Ctext%7B%20of%20ace%3D%201%7D%20%5Cend%7Bgathered%7D)
Thus, we have the probability to be evaluated as
Answer:
<em>The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938
</em>
Step-by-step explanation:
<em>Given:</em>
<em>Here Section 1 students = 20
</em>
<em>
Section 2 students = 30
</em>
<em>
Here there are 15 graded exam papers.
</em>
<em>
(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070
</em>
<em>
(b) Here if x is the number of students copies of section 2 out of 15 exam papers.
</em>
<em> here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15
</em>
<em>Then,
</em>
<em>
Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798
</em>
<em>
(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)
</em>
<em>
so,
</em>
<em>
Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938
</em>
<em>
Note : Here the given distribution is Hyper-geometric distribution
</em>
<em>
where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>
Answer: length is 10 feet, width is 6.5 feet
Step-by-step explanation: