Question

find the probability, to 4 decimal places:
it is possible when rounded that a probability is 0.0000
d) exactly none are left - handed.

e) exactly 10 are left - handed.

f) at least 4 are left - handed.

g) at most 3 are left - handed.

h) at least 6 are left - handed.

Answer:

To solve these probability problems, we assume a binomial probability distribution (since we're dealing with the number of left - handed people, a binary outcome: left - handed or not left - handed). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successful trials, \(p\) is the probability of success on a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\) is the number of combinations of \(n\) items taken \(k\) at a time. However, we need to know the values of \(n\) (the total number of people) and \(p\) (the probability that a person is left - handed) to proceed. Since these values are not provided in the problem, we cannot calculate the probabilities.

If we assume some common values (for example, if \(n = 10\) and \(p=0.1\) (approximate probability of being left - handed)):

Part d: Exactly none are left - handed (\(k = 0\))

Step 1: Identify values

Let \(n\) be the number of people, \(k = 0\), \(p\) be the probability of being left - handed, and \(1-p\) be the probability of being right - handed.

Step 2: Apply binomial formula

\(P(X = 0)=C(n,0)\times p^{0}\times(1 - p)^{n-0}\)
Since \(C(n,0) = 1\) and \(p^{0}=1\), \(P(X = 0)=(1 - p)^{n}\)

If \(n = 10\) and \(p = 0.1\), then \(P(X=0)=(1 - 0.1)^{10}=0.9^{10}\approx0.3487\)

Part e: Exactly 10 are left - handed (\(k = 10\))

Step 1: Identify values

\(n\) is the number of people, \(k = 10\), \(p\) is the probability of being left - handed.

Step 2: Apply binomial formula

\(P(X = 10)=C(n,10)\times p^{10}\times(1 - p)^{n - 10}\)
If \(n = 10\) and \(p=0.1\), \(C(10,10) = 1\), \(P(X = 10)=1\times(0.1)^{10}\times(0.9)^{0}=1\times10^{- 10}\times1\approx0.0000\)

Part f: At least 4 are left - handed (\(P(X\geq4)=1 - P(X\lt4)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3))\))

Step 1: Calculate \(P(X = 0)\), \(P(X = 1)\), \(P(X = 2)\), \(P(X = 3)\)

Using the binomial formula \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\)
If \(n = 10\) and \(p = 0.1\):

• \(P(X = 0)=(0.9)^{10}\approx0.3487\)
• \(P(X = 1)=C(10,1)\times0.1^{1}\times0.9^{9}=10\times0.1\times0.3874\approx0.3874\)
• \(P(X = 2)=C(10,2)\times0.1^{2}\times0.9^{8}=\frac{10!}{2!(10 - 2)!}\times0.01\times0.4305 = 45\times0.01\times0.4305\approx0.1937\)
• \(P(X = 3)=C(10,3)\times0.1^{3}\times0.9^{7}=\frac{10!}{3!(10 - 3)!}\times0.001\times0.4783=120\times0.001\times0.4783\approx0.0574\)

Step 2: Calculate \(P(X\lt4)\)

\(P(X\lt4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)=0.3487 + 0.3874+0.1937 + 0.0574=0.9872\)

Step 3: Calculate \(P(X\geq4)\)

\(P(X\geq4)=1 - 0.9872 = 0.0128\)

Part g: At most 3 are left - handed (\(P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\))

Using the values calculated in part f, \(P(X\leq3)=0.3487+0.3874 + 0.1937+0.0574 = 0.9872\)

Part h: At least 6 are left - handed (\(P(X\geq6)=P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)\))

Step 1: Calculate individual probabilities

• \(P(X = 6)=C(10,6)\times0.1^{6}\times0.9^{4}=\frac{10!}{6!(10 - 6)!}\times10^{-6}\times0.6561=210\times10^{-6}\times0.6561\approx0.0001\)
• \(P(X = 7)=C(10,7)\times0.1^{7}\times0.9^{3}=\frac{10!}{7!(10 - 7)!}\times10^{-7}\times0.729 = 120\times10^{-7}\times0.729\approx0.0000\)
• \(P(X = 8)=C(10,8)\times0.1^{8}\times0.9^{2}=\frac{10!}{8!(10 - 8)!}\times10^{-8}\times0.81=45\times10^{-8}\times0.81\approx0.0000\)
• \(P(X = 9)=C(10,9)\times0.1^{9}\times0.9^{1}=10\times10^{-9}\times0.9\approx0.0000\)
• \(P(X = 10)=0.0000\) (from part e)

Step 2: Sum the probabilities

\(P(X\geq6)=0.0001+0.0000 + 0.0000+0.0000+0.0000=0.0001\)

But since the values of \(n\) and \(p\) are not given in the original problem, the above calculations are based on assumptions. To get the correct answers, we need to know the values of \(n\) (total number of people) and \(p\) (probability that a person is left - handed).

If you can provide the values of \(n\) (the number of people in the sample) and \(p\) (the probability that a single person is left - handed), we can give you the exact probabilities.