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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 ?

1. $\left(\dfrac{7}{8}\right)$
2. $\left(\dfrac{1}{2}\right)$
3. $\left(\dfrac{7}{16}\right)$
4. $\left(\dfrac{5}{32}\right)$

Answer is C) $\dfrac{7}{16}$

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

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

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

by using rule of total probility

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

### 1 comment

cleanest ever method
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)
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