The probability of Andy and Anna having lunch together is 40%. The probability of them having lunch together at least 6 days a w
1 answer:
The probability of having lunch together is p = 40% = 0.4
The probability of not having lunch together is q = 1 - p = 0.6
Number of trials (days in a week) is n = 7
Let r = number of days in the week when Andy and Anna have lunch together.
Use th graphing calculator to obtain
P(6 of 7) = ₇C₆ (0.4)⁶(0.6) = 0.017
P(7 of 7) = ₇C₇ (0.4)⁷(0.6)⁰ = 0.002
Therefore
P(at least 6 of 7) = P(1 of 7) + P(2 of 7) + ... + P(6 of 7)
= 0.131 + 0.261 + 0.290 + 0.194 + 0.077 + 0.017
= 0.97 or 97%
P(at least 6 of 7) = 0.017 + 0.002 = 0.019 = 1.9%
P(exactly 6 of 7) = 0.017 or 1.7%
Answer:
The probability of having lunch at least 6 days per week is 0.019 or 1.9%.
The probability of having lunch exactly 6 times is 0.017 or 1.7%
The probability of not having lunch together is q = 1 - p = 0.6
Number of trials (days in a week) is n = 7
Let r = number of days in the week when Andy and Anna have lunch together.
Use th graphing calculator to obtain
P(6 of 7) = ₇C₆ (0.4)⁶(0.6) = 0.017
P(7 of 7) = ₇C₇ (0.4)⁷(0.6)⁰ = 0.002
Therefore
P(at least 6 of 7) = P(1 of 7) + P(2 of 7) + ... + P(6 of 7)
= 0.131 + 0.261 + 0.290 + 0.194 + 0.077 + 0.017
= 0.97 or 97%
P(at least 6 of 7) = 0.017 + 0.002 = 0.019 = 1.9%
P(exactly 6 of 7) = 0.017 or 1.7%
Answer:
The probability of having lunch at least 6 days per week is 0.019 or 1.9%.
The probability of having lunch exactly 6 times is 0.017 or 1.7%
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Answer:
E(X₁)= 1.76
E(X₂)= 4.18
E(X₃)= 9.24
E(X₄)= 6.82
a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022
b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001
c. P(X₁≤2) = 0.7442
Step-by-step explanation:
Hello!
So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:
X₁: "The seedling is dead" p₁: 0.08
X₂: "The seedling exhibits slow growth" p₂: 0.19
X₃: "The seedling exhibits medium growth" p₃: 0.42
X₄: "The seedling exhibits strong growth" p₄:0.31
To calculate the expected number for each category (k) you need to use the formula:
E(X
So
E(X₁)= n*p₁ = 22*0.08 = 1.76
E(X₂)= n*p₂ = 22*0.19 = 4.18
E(X₃)= n*p₃ = 22*0.42 = 9.24
E(X₄)= n*p₄ = 22*0.31 = 6.82
Next, to calculate each probability you just use the corresponding probability of success of each category:
Formula: P(X₁, X₂,..., Xk) =
a.
P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= = 0.00022
b.
P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= = 0.000001
c.
P(X₁≤2) = +
+
= 0.7442
I hope you have a SUPER day!