Probability of Head, P(H) = 0.6
Probability of Tail, P(T) = 0.4

Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.6)^3*(0.4)^2
1 Tail and 4 Heads, Probability = 5C1*(0.6)^4*(0.4)^2
0 Tail and 5 Heads, Probability = (0.6)^5

Required Probability = Sum of all Favourable cases = (0.6)^5 + 5(0.6)^4(0.4) + 10(0.6)^3(0.4)^2

Given that A coin is weighted so that the probability of heads on any flip is 0.6, while the probability of tails is 0.4. If the coin is flipped 5 times independently. We need to find which of the following represents the probability that tails will appear no more than twice?

As the coin is tossed five times then number of cases = \(2^5\) = 32

P(Tails will appear no more than twice) = P(0T) + P(1T) + P(2T)

P(0T)

P(0T) = P(HHHHH) = 0.6*0.6*0.6*0.6*0.6 = \((0.6)^5\)

P(1T)

Now out of the five place _ _ _ _ _ Tail can come in any of the places in 5C1 ways = 5 ways

=> P(1T) = Number of places * P(Tail) * P(Head) * P(Head) * P(Head) * P(Head) = 5*0.4*0.6*0.6*0.6*0.6 = \(5(0.6)^4(0.4)\)

P(2T)

Now out of the five place _ _ _ _ _ we need to find two places where Tail can come => 5C2 ways = \(\frac{5!}{2!*(5-2)!}\) ways = \(\frac{5*4*3!}{2!*3!}\) ways = 10 ways

=> P(2T) = Number of places * P(Tail) * P(Tail) * P(Head) * P(Head) * P(Head) = 10*0.4*0.4*0.6*0.6*0.6 = \(10(0.6)^3(0.4)^2\)

P(Tails will appear no more than twice) = P(0T) + P(1T) + P(2T) = \((0.6)^5\) + \(5(0.6)^4(0.4)\) + \(10(0.6)^3(0.4)^2\)

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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