This is due today I don’t get it.
2 answers:
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Tossing a coin is a binomial experiment.
Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.
All of these trials are independent since the result of one trial does not affect the result of the next trial.
Now, for 'n' repeated trials the total number of successes is given by
where 'r' denotes the number of successful results.
In our case and
,
Substituting the values we get,
Therefore, there are 1352078 ways to get heads if a person tosses a coin 23 times.
Answer:
Step-by-step explanation:
Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by
f(x,y) = P((X=x,Y=y) ,
and given that the random variables are independent the joint pmf isbgiven by:
f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)
(b) the required probability is given by considering X=0,1 and Y =0,1
f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }
= [0.1 +0.2 ] × [0.1 + 0.3]
= 0.3 × 0.4
= 0.12
Now we find
fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1
fX(x) = sum{[f(x,y)]}= 0.1+0.3
= 0.4
fY(y) = sum{[f(x.y)]} = 0.1 + 0.2
= 0.3
Since fY(y) × fX(x) = 0.4 × 0.3
= 0.12
Hence f(x,y) = fX(x) .fY(y), the events are independent
(c) the required event is give by: [P(X<=1, PY<=1)]
= P(X<=1) . P(Y<=1)
= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1
= 0.4 × 0.3
= 0.12
(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A
A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]
A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]
A = [0.09] + [ 0.09]
A = 0.18
Pls note the sum of the vertical column X=2 is 0.3
