Question 1082101: A coin is biased such that the head is twice as likely to occur as tail. If this coin is tossed 4 times, find
i- the sample space
ii- the probability of having at least 2 heads
iii- the probability of having exactly 3 heads
iv- the probability of having at most 4 heads.
Found 2 solutions by Boreal, rothauserc:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
HTTH
THHT
THTH
TTHH
TTTH
TTHT
THTT
HTTT
TTTT
Exactly two heads in 4 tosses: 6 ways of having them 4C2
The denominator of the product of the probabilities is always 81. The numerator numbers change in order but not in product, and they are 2*2*1*1=4. Multiply by 6 and get 24/81=8/27
exactly 3 heads is 4 * 2*2*2*1/87=32/81
exactly 4 heads is 2*2*2*2/81=16/81
(and 1 head and 0 heads for completeness:
1:2*4/81=8/81
0:1/81
Sum is 81/81
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! Let Probability(P) of getting a tail on a toss = x
:
Let P of getting a head on a toss = 2x
:
we know the sum of the probabilities = 1
:
x + 2x = 1
:
x = 1/3
:
P of getting a tail on a toss = 1/3
P of getting a head on a toss = 2/3
:
i) there are 2^4 possible combinations
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,
THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
:
Note that there is no difference between tossing 1 coin 4 times and tossing
4 coins once
:
We use the binomial probability for ii and iv
:
ii) P(at least 2 heads) = 1 - P(0 heads) - P(1 head)
Note P(k successes in n trials) = nCk * p^k * (1-p)^n-k, where p = 2/3, n = 4,
4Ck = 4! / (k! * (n-k)!
P(at least 2 heads) = 1 - 0.01 - 0.10 = 0.89
:
iii) P(of 3 heads) = 0.40
:
iv) P(at most 4 heads) = 1 - P(0 heads)
P(at most 4 heads) = 1 - 0.01 = 0.99
:
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