SOLUTION: A coin is tossed five times. Find the probability that the sixth toss is a tail, given that the first five tosses were all tails. If a die is rolled one time, find the probabili

Question 349911: A coin is tossed five times. Find the probability that the sixth toss is a tail, given that the first five tosses were all tails.
If a die is rolled one time, find the probability of getting a number greater than five.
If one card is drawn from an ordinary deck of cards, what is the probability that the card will be an ace, a king of hearts, or a spade.
A pet supplier has a stock of parkeets of which thirty percent are blue parakeets. A pet store orders three parakeets from this supplier. If supplier selects the parakeets at random, what is the chance that the pet store gets exactly one blue parakeet?
A university has 10,000 students of which 4000 are male and 6000 are female. If a class of 30 students is chosen at random from university population, find the mean and variance of the number of male students.

Found 2 solutions by solver91311, Fombitz:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

One question per post.

Find the probability that the sixth toss is a tail, given that the first five tosses were all tails

Presuming that the coin is a fair coin, the probability that the sixth toss is a tail is the same as any other trial, namely . But since the probability of tossing 5 tails in a row with a fair coin is only , you might want to investigate whether the coin that tossed 5 tails in a row was actually a fair coin.

John

My calculator said it, I believe it, that settles it

Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
I'll do 4 of the problems.
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a) P(T)=P(H)=1/2, it's independent of previous tosses if the coin is fair.
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b) P(6)=1/6
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c) P(Ace)=4/52=1/13
P(King of Hearts)=1/52
P(Spade)=4/52=1/13
P(Ace or K Hearts or Spade)=P(Ace)+P(K Hearts)+P(Spade)-P(Ace of Spades)
So we don't double count the Ace of Spades as an Ace and a Spade.
P(Ace of Spades)=1/52
P(Ace or K Hearts or Spade)=4/52+1/52+13/52-1/52=17/52
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d) P(B,NB,NB)=0.3(0.7)(0.7)
P(NB,B,NB)=0.7(0.3)(0.7)
P(NB,NB,B)=0.7(0.7)(0.3)
P(1 B)=3(0.3)(0.7)(0.7)=0.441
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