GATE IT 2007 | Question: 1

Question

Suppose there are two coins. The first coin gives heads with probability $\dfrac{5}{8}$ when tossed, while the second coin gives heads with probability $\dfrac{1}{4}.$ One of the two coins is picked up at random with equal probability and tossed. What is the probability of obtaining heads ?

• A. $\left(\dfrac{7}{8}\right)$
• B. $\left(\dfrac{1}{2}\right)$
• C. $\left(\dfrac{7}{16}\right)$
• D. $\left(\dfrac{5}{32}\right)$

Comments

Here the confusion can be created where we thought already in the mind the We should take First C1 and then C2 or C2 then C1 but it was not asked!
 One of the two coins is picked up at random with equal probability and tossed.
we have the choice of picking the one of the coin

Answer 1

Answer is C) $\dfrac{7}{16}$

Probability of obtaining head$= \text{Probability of picking first coin} \times \text{Probability of getting head on first coin}$

$+ \text{Probability of picking second coin}\times \text{Probability of getting head on second coin}$

$ = \left(\dfrac{1}{2}\times \dfrac{5}{8}\right) +\left( \dfrac{1}{2}\times \dfrac{1}{4}\right) = \dfrac{7}{16}.$

Answer 2

by using rule of total probility

p(head) =(1/2)*(5/8) + (1/2)*(1/4) = 7/16 

Comments

cleanest ever method

Answer 3

 the answer is Option C

In the calculation, I made a mistake P(head| Coin1) =5/8, I mistakenly write 3/8.

Answer 4

by total probability ,

                           C1                C2

For Heads :       5/8               ¼

 

equal

probability   :       ½                ½

 

                          Total Probability(for heads) = 5/8 * ½  +  ¼ * ½  = 7/16     -ANS (C)