Question

Mark the correct alternative in each of the following: Two different coins are tossed simultaneously. The probability of getting at least one head is
(a)$\frac{1}{4}$
(b)$\frac{1}{8}$
(c)$\frac{3}{4}$
(d)$\frac{7}{8}$

Answer 1

SOLUTION :  

The correct option is (C)   : ¾  

Given : Two coins are tossed simultaneously  

If two coins are tossed then the possible outcomes are : HT, HH, TT, TH

Total no. of possible outcomes = 4

Let E = Event of getting at least one head

Outcomes favourable to  E = HT, TH, HH

No. of favorable outcomes = 3

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 3/4  

Hence, the required probability of getting at least one head ,P(E) = 3/4 .

HOPE THIS ANSWER WILL HELP  YOU…

Answer 2

✌️✌️ Your answer ✌️✌️

✔️✔️ OPTION C✔️✔️

✳️ Explaination ✳️

⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️

$\large = >probablity = \frac{no.of \: favourable \: outcomes}{no.of \: total \: outcomes}$

=> Head - Head

=> Tail - Head

=> Tail - Head

=> total outcomes: 4

=> possible outcomes: 3

=> 3/4

=> Therefore the answer is

✔️✔️option C ✔️✔️

✔️✔️3/4✔️✔️