IIT JEE , AIEEE Forum - Mathematics , Algebra

mujtaba4iit

A coin is tossed 4times. Find the probability that there is no tail in the results.

ziauddin75

in i toss
p=1/2
q=1/2
n=4
P=4C0 (p)^4 q^0

ans. is 1/16

if right then rate me

sboosy

P (Head)= 1/2 (assuming fair coin)=P(tail)

all we want is a head on all the occassions

the probablity of which is 1/2 * 1/2 * 1/2 * 1/2 = 1/16

rishipratimm

the ans is....
(1/4)^4
= 1/16....

mujtaba4iit

and what if:
for the same condition find the probability of getting atleast one tail.
Plz state the sample space for the condition.

sboosy

it is 1- probability of getting no tail
so 1- 1/16 = 15/16

sboosy

explanation :
see in tossing a coin 4 times
u could get 0 or 1 or 2 or 3 or all 4 as tails ......1
atleast 1 tail means either 1 or 2 or 3 or 4
that is expecting 0
now since 1 is exhaustive sum of all probabilities is 1
so it is 1- prob of getting no tail

rishipratimm

ooopsie....stupid slip....
it shud hav been (1/2)^4....
sorry 4 the mstk.....

ziauddin75

P(getting atleast one tail)=l-P(getting no tail)

so
1-1/16(fromm my prev reply)
15/16

mujtaba4iit

yes rishi i understood the slip. I gave u a 2 pointer/

Thanx evry1 else too.

chetan_kp

using the laws of probability the no. of ways in which tail occurs in all the results is same as getting all heads
no. of ways of getting all heads = (1/2)*(1/2)*(1/2)*(1/2)
= 1/16

chetan_kp

for the second part u have mentioned :
sample space = ²C₁*²C₁*²C₁*²C₁
= 2*2*2*2
= 16

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