Latest TCS Aptitude Question SOLUTION: A die is rolled and a coin is tossed. Find the probability that the die shows an odd number and the coin shows a head.

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(#M40018507) TCS Numerical Ability question Keep an EYE

A die is rolled and a coin is tossed. Find the probability that the die shows an odd number and the coin shows a head.

Asked In TCS Rajib Chakraborty (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

amrit  (10 years ago)

The sample space S of the experiment described in question 5 is as follows

S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)
(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}
Let E be the event "the die shows an odd number and the coin shows a head". Event E may be described as follows
E={(1,H),(3,H),(5,H)}
# The probability P(E) is given by
P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

Is this solution Helpfull? Yes (31) | No (1)  

vinod  (10 years ago)

they are two independent events, p(a&b)=p(a)*p(b) when a&b are two independent events
here p(a)= odd number prob= 3/6=1/2
p(b)= coin show head= 1/2
so ans= 1/2 * 1/2 = 1/4

Is this solution Helpfull? Yes (12) | No   

devendra  (10 years ago)

1/4
even tell
even head
odd tell
odd head
probability=1/4

Is this solution Helpfull? Yes (4) | No (2)  

Varun Shivani  (10 years ago)

for coin 3/6=1/2 and for a head prob will be 1/2 .
so 1/2 *1/2=1/4 is the answer.

Is this solution Helpfull? Yes (2) | No   

Jasmine Khan  (10 years ago)

s={(1,H)(2,H)(3,H)(4,H)(5,H)(6,H)(1,T)(2,T)(3,T)(4,T)(5,T)(6,T)}
no. of favorable outcomes=3
total no. of outcomes=12
p(the die shows odd number and the coin shows a head)=3/12=1/4

Is this solution Helpfull? Yes (1) | No   

Ann Mary Jolly  (10 years ago)

ANS : 0.25
No. of favourable outcomes = 3 [since there are 3 odd no.s and one head]
Total no. of outcomes = 6*2 = 12
Therefore required probability = no. of favourable outcomes/total outcomes
= 3/12 = 1/4 = .25

Is this solution Helpfull? Yes (1) | No   

neelima datla  (10 years ago)

probability for die to show odd no. is 3/6=1/2
and for coin to show head is 1/2.
probability that die shows odd no. and coin a head is 1/2+1/2=1
by the formula p(aub)=p(a)+p(b)

Is this solution Helpfull? Yes (1) | No   

monika  (10 years ago)

s={(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}
no.of favourable outcomes=3
total no.of outcomes=12
probability=3/12=1/4

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KARTHIGA DEVI.J  (10 years ago)

for coin 3/6=1/2 and for a head prob will be 1/2 .
'and'means multiply,so
1/2 * 1/2 = 1/4

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nimisha jain  (10 years ago)

p(odd)=1/6
pair(1,t),(3,t),(5,t)
p(t)=1/2
p(0ddnumber&coin )= 1/6*1/2+1/6*1/2
1/6*1/2
3/12=1/4
ans=1/4

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akhilesh sreelayam  (10 years ago)

6C3(combinations of odd no. from dice)*2C1(combination of heads from coin)..

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raushan kumar  (10 years ago)

(1/2)*(3/6)= 1/4

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Diwakar Kumar  (10 years ago)

Let e be the event of getting a odd number from rolling a dice.
therefore e={1,3,5},ans s be the sample space which is s={1,2,3,4,5,6}
hence p(e)=3/6=1/2.
and p(Tail)=1/2.
as both connected with and operation so. total probability =1/2*1/2=1/4.

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kriti  (10 years ago)

sample space of throwing a die=6
sample space of tossing a coin =2
therefore event of getting an odd no and a head=3/6*1/2=1/4

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raviteja  (10 years ago)

a die is rolled then tne probability that the die shows an odd number=3/6
coin shows head=1/2
he probability that the die shows an odd number and the coin shows a head=3/6*1/2=1/4

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