Question s +o.5 points). Suppose that on each day, the probability of Donald Tru
ID: 3316175 • Letter: Q
Question
Question s +o.5 points). Suppose that on each day, the probability of Donald Trump insulting some- one on Twitter is 20%. What is the probability that, during the week from December 11 to 17, 2017, there will be at least three consecutive days during which he does not insult anyone on Twitter? Question 6 (+1 points). The Pie family has two children, who are equally likely to be boys or girls. The chance of a girl in the Pie family being named "Apple" is I in 1oo. (This is a constant; even if there is already an Apple Pie, the chance of the next Pie-if a girl-to be named Apple is still in 1o0.) Whats the probability of both children being girls, given that at least one of them is named Apple?Explanation / Answer
Answer to question# 5)
P(insulting) = 0.20
Total number of days = 7
11th 12th 13th , or 12th , 13th , 14th or 13th , 14th , 15th , or 14th , 15th 16th , or 15th , 16th ,17th
in this way 3 consecutie days can be formed
Thus we get 5 alternatives
P(of first option) = (0.20)^3*(0.80)^4 = 0.0032768
[that is 3 days he insults, and rest of the four days he doesnot insult]
P(3 consecutive days) = 5*0.0032768 = 0.016384
.
Like thos there would be 4 alternatives for 4 consecutive days
P(4 days) = 4*(0.20)^4 * (0.80)^3 = 0.0032768
.
Like this for 5 days , there are 3 options
P(5 days) = 3* (0.2)^5 * (0.8)^3 = 0.000492
.
Like this for 6 days there are 2 options
P(6 days) = 6 *0.2^6 *0.8 = 0.000307
.
For 7 days , it would be all 7 days
P(7 days) = 0.2^7 = 0.0000128
.
Thus total probability = P(3 days) + P(4 days) + P( 5days) + P(6 days) + P(7 days)
Total probability = 0.016384 + 0.0032768 + 0.000492 + 0.0003072 + 0.0000128 = 0.02047
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